From 10e26ba7459520274dc1816c13c060a449174570 Mon Sep 17 00:00:00 2001 From: Cory Wilson <40644848+cg2wilson@users.noreply.github.com> Date: Thu, 22 May 2025 18:19:35 +0000 Subject: [PATCH 1/4] add answers to TI1 --- source/calculus/source/05-TI/01.ptx | 445 ++++++++++++++++++---------- 1 file changed, 290 insertions(+), 155 deletions(-) diff --git a/source/calculus/source/05-TI/01.ptx b/source/calculus/source/05-TI/01.ptx index 7484401cc..ef4776071 100644 --- a/source/calculus/source/05-TI/01.ptx +++ b/source/calculus/source/05-TI/01.ptx @@ -23,6 +23,11 @@ Using the chain rule, which of these is the derivative of e^{x^3} with re

  • 3x^2e^{x^3}

  • \dfrac{1}{4}e^{x^4}

    • + +

    + C: 3x^2e^{x^3} +

    +

    @@ -34,6 +39,11 @@ Based on this result, which of these would you suspect to equal \displaystyle

  • 3e^{x^3}+C

  • \dfrac{1}{3}e^{x^3}+C

    • + +

    + D: \dfrac{1}{3}e^{x^3}+C +

    +     @@ -52,16 +62,31 @@ Based on this result, which of these would you suspect to equal \displaystyle

    What is \displaystyle \int 7u^6 u'\,dx?

     + +

    + B: u^7 + C +

    +

    What is \displaystyle \int u^6 u'\,dx?

     + +

    + A: \dfrac{1}{7}u^7 + C +

    +

    What is \displaystyle \int 6u^6 u'\,dx?

     + +

    + D: \dfrac{6}{7}u^7 + C +

    +     @@ -73,15 +98,19 @@ Based on this result, which of these would you suspect to equal \displaystyle

  • Keep differentiating functions until you come across the function you want to integrate.

    • + +

    + B: Attempt to rewrite the integral in the form \displaystyle \int g'(u)u'\,dx=g(u)+C. +

    +

    -By the chain rule, \dfrac{d}{dx}[g(u)+C]=g'(u)u'. There is a dual integration technique reversing this process, known as the substitution method. + By the chain rule, \dfrac{d}{dx}[g(u)+C]=g'(u)u'. There is a dual integration technique reversing this process, known as the substitution method.

    -This technique involves choosing an appropriate function u in terms of x -to rewrite the integral as follows: + This technique involves choosing an appropriate function u in terms of x to rewrite the integral as follows: \displaystyle \int f(x)\,dx=\dots=\displaystyle \int g'(u)u'\,dx=g(u)+C.

    @@ -89,188 +118,194 @@ to rewrite the integral as follows:

    -Recall that \dfrac{du}{dx}=u', and so du=u'\,dx. This allows for the following common notation: + Recall that \dfrac{du}{dx}=u', and so du=u'\,dx. This allows for the following common notation: \displaystyle \int f(x)\,dx=\dots=\displaystyle \int g'(u)\, du=g(u)+C.

    -Therefore, rather than dealing with equations like u'=\dfrac{du}{dx}=x^2, we will prefer to write -du=x^2\, dx. + Therefore, rather than dealing with equations like u'=\dfrac{du}{dx}=x^2, we will prefer to write du=x^2\, dx.

    -Consider \displaystyle \int x^2e^{x^3}\,dx, which we conjectured earlier -to be \dfrac{1}{3}e^{x^3}+C. + Consider \displaystyle \int x^2e^{x^3}\,dx, which we conjectured earlier to be \dfrac{1}{3}e^{x^3}+C.

    -Suppose we decided to let u=x^3. + Suppose we decided to let u=x^3.

    -Compute \dfrac{du}{dx}=\unknown, -and rewrite it as du=\unknown\,dx. + Compute \dfrac{du}{dx}=\unknown, and rewrite it as du=\unknown\,dx.

    + +

    + \dfrac{du}{dx} = 3u^2, so du = 3u^2\, dx +

    +

    -This \unknown\,dx doesn't appear in \displaystyle \int x^2e^{x^3}\,dx -exactly, so use algebra to solve for x^2\,dx in terms of du. + This \unknown\,dx doesn't appear in \displaystyle \int x^2e^{x^3}\,dx exactly, so use algebra to solve for x^2\,dx in terms of du.

    + +

    + \dfrac{1}{3}\, du = x^2\, dx +

    +

    -Replace x^2\, dx and x^3 with u\,du terms to rewrite -\displaystyle \int x^2e^{x^3}\,dx as -\displaystyle \int \dfrac{1}{3}e^u\,du. + Replace x^2\, dx and x^3 with u\,du terms to rewrite \displaystyle \int x^2e^{x^3}\,dx as \displaystyle \int \dfrac{1}{3}e^u\,du.

    -Solve \displaystyle \int \dfrac{1}{3}e^u\,du in terms of u, -then replace u with x^3 to confirm our original -conjecture. + Solve \displaystyle \int \dfrac{1}{3}e^u\,du in terms of u, then replace u with x^3 to confirm our original conjecture.

    + +

    + \displaystyle \int \dfrac{1}{3}e^u\, du = \dfrac{1}{3}e^u + C = \dfrac{1}{3}e^{x^3} + C +

    +

    -Here is how one might write out the explanation -of how to find -\displaystyle \int x^2e^{x^3}\,dx from start to finish: + Here is how one might write out the explanation of how to find \displaystyle \int x^2e^{x^3}\,dx from start to finish:

    -\displaystyle \int x^2e^{x^3}\,dx &&\text{Let }&u=x^3 + \displaystyle \int x^2e^{x^3}\,dx &&\text{Let }&u=x^3 -&&& du = 3x^2\,dx + &&& du = 3x^2\,dx -&&& \dfrac{1}{3}du = x^2\,dx + &&& \dfrac{1}{3}du = x^2\,dx -\displaystyle \int x^2e^{x^3}\,dx &= \displaystyle \int e^{(x^3)} (x^2\,dx) + \displaystyle \int x^2e^{x^3}\,dx &= \displaystyle \int e^{(x^3)} (x^2\,dx) -&= \displaystyle \int e^{u} \dfrac{1}{3}\,du + &= \displaystyle \int e^{u} \dfrac{1}{3}\,du -&= \dfrac{1}{3}e^{u}+C + &= \dfrac{1}{3}e^{u}+C -&= \dfrac{1}{3}e^{x^3}+C + &= \dfrac{1}{3}e^{x^3}+C     +

    -Which step of the previous example do you think was -the most important? + Which step of the previous example do you think was the most important?

    1. -Choosing u=x^3. + Choosing u=x^3.

    2. -Finding du=3x^2\,dx and \dfrac{1}{3}du=x^2\,dx. + Finding du=3x^2\,dx and \dfrac{1}{3}du=x^2\,dx.

    3. -Substituting \displaystyle \int x^2e^{x^3}\,dx -with \displaystyle \int\dfrac{1}{3}e^u\,du. + Substituting \displaystyle \int x^2e^{x^3}\,dx with \displaystyle \int\dfrac{1}{3}e^u\,du.

    4. -Integrating \displaystyle \int\dfrac{1}{3}e^u\,du=\dfrac{1}{3}e^u+C. + Integrating \displaystyle \int\dfrac{1}{3}e^u\,du=\dfrac{1}{3}e^u+C.

    5. -Unsubstituting \dfrac{1}{3}e^u+C -to get \dfrac{1}{3}e^{x^3}+C. + Unsubstituting \dfrac{1}{3}e^u+C to get \dfrac{1}{3}e^{x^3}+C.

    + +

    + Answers might vary, as the question asks the student's opinion. +

    +

    -Below are two correct solutions to the same integral, using -two different choices for u. Which method would you prefer -to use yourself? + Below are two correct solutions to the same integral, using two different choices for u. Which method would you prefer to use yourself?

    -\displaystyle \int x\sqrt{4x+4}\,dx &\phantom{=}\text{Let }u=x+1 + \displaystyle \int x\sqrt{4x+4}\,dx &\phantom{=}\text{Let }u=x+1 -&\phantom{=} 4u=4x+4 + &\phantom{=} 4u=4x+4 -&\phantom{=} x=u-1 + &\phantom{=} x=u-1 -&\phantom{=} du = dx + &\phantom{=} du = dx -\displaystyle \int x\sqrt{4x+4}\,dx &= \displaystyle \int (u-1)\sqrt{4u}\,du + \displaystyle \int x\sqrt{4x+4}\,dx &= \displaystyle \int (u-1)\sqrt{4u}\,du -&= \displaystyle \int (2u^{3/2}-2u^{1/2})\,du + &= \displaystyle \int (2u^{3/2}-2u^{1/2})\,du -&= \dfrac{4}{5}u^{5/2}-\dfrac{4}{3}u^{3/2}+C + &= \dfrac{4}{5}u^{5/2}-\dfrac{4}{3}u^{3/2}+C -&= \dfrac{4}{5}(x+1)^{5/2} + &= \dfrac{4}{5}(x+1)^{5/2} -&\phantom{=}-\dfrac{4}{3}(x+1)^{3/2}+C + &\phantom{=}-\dfrac{4}{3}(x+1)^{3/2}+C

    -\displaystyle \int x\sqrt{4x+4}\,dx &\phantom{=}\text{Let }u=\sqrt{4x+4} + \displaystyle \int x\sqrt{4x+4}\,dx &\phantom{=}\text{Let }u=\sqrt{4x+4} -&\phantom{=} u^2=4x+4 + &\phantom{=} u^2=4x+4 -&\phantom{=} x=\dfrac{1}{4}u^2-1 + &\phantom{=} x=\dfrac{1}{4}u^2-1 -&\phantom{=} dx=\dfrac{1}{2}u\,du + &\phantom{=} dx=\dfrac{1}{2}u\,du -\displaystyle \int x\sqrt{4x+4}\,dx &= \displaystyle \int \left(\dfrac{1}{4}u^2-1\right)(u)\left(\dfrac{1}{2}u\,du\right) + \displaystyle \int x\sqrt{4x+4}\,dx &= \displaystyle \int \left(\dfrac{1}{4}u^2-1\right)(u)\left(\dfrac{1}{2}u\,du\right) -&= \displaystyle \int \left(\dfrac{1}{8}u^4-\dfrac{1}{2}u^2\right)\,du + &= \displaystyle \int \left(\dfrac{1}{8}u^4-\dfrac{1}{2}u^2\right)\,du -&= \dfrac{1}{40}u^5-\dfrac{1}{6}u^3+C + &= \dfrac{1}{40}u^5-\dfrac{1}{6}u^3+C -&= \dfrac{1}{40}(4x+4)^{5/2} + &= \dfrac{1}{40}(4x+4)^{5/2} -&\phantom{=}-\dfrac{1}{6}(4x+4)^{3/2}+C + &\phantom{=}-\dfrac{1}{6}(4x+4)^{3/2}+C

    @@ -280,99 +315,139 @@ to use yourself?

    -Suppose we wanted to try the substitution method to find \displaystyle \int e^x\cos(e^x+3)\,dx. Which of these choices for u appears to be most useful? + Suppose we wanted to try the substitution method to find \displaystyle \int e^x\cos(e^x+3)\,dx. Which of these choices for u appears to be most useful?

    1. -u=x, so du=dx + u=x, so du=dx

    2. -u=e^x, so du=e^x\,dx + u=e^x, so du=e^x\,dx

    3. -u=e^x+3, so du=e^x\,dx + u=e^x+3, so du=e^x\,dx

    4. -u=\cos(x), so du=-\sin(x)\,dx + u=\cos(x), so du=-\sin(x)\,dx

    5. -u=\cos(e^x+3), so du=e^x\sin(e^x+3)\,dx + u=\cos(e^x+3), so du=e^x\sin(e^x+3)\,dx

    + +

    + C: u=e^x+3, so du=e^x\,dx +

    +

    -Complete the following solution using your choice from -the previous activity to find \displaystyle \int e^x\cos(e^x+3)\,dx. + Complete the following solution using your choice from the previous activity to find \displaystyle \int e^x\cos(e^x+3)\,dx.

    -\displaystyle \int e^x\cos(e^x+3)\,dx &&\text{Let }&u=\unknown + \displaystyle \int e^x\cos(e^x+3)\,dx &&\text{Let }&u=\unknown -&&& du = \unknown\,dx + &&& du = \unknown\,dx -\displaystyle \int e^x\cos(e^x+3)\,dx &= \displaystyle \int \unknown\, du + \displaystyle \int e^x\cos(e^x+3)\,dx &= \displaystyle \int \unknown\, du -&= \cdots + &= \cdots -&= \sin(e^x+3)+C + &= \sin(e^x+3)+C     + + + + \displaystyle \int e^x\cos(e^x+3)\,dx &&\text{Let }&u=e^x + 3 + + + &&& du = e^x\,dx + + + \displaystyle \int e^x\cos(e^x+3)\,dx &= \displaystyle \int \cos(u)\, du + + + &= \sin(u) + C + + + &= \sin(e^x+3)+C + + +

    -Complete the following integration by substitution -to find \displaystyle \int \dfrac{x^3}{x^4+4}\,dx. + Complete the following integration by substitution to find \displaystyle \int \dfrac{x^3}{x^4+4}\,dx.

    -\displaystyle \int \dfrac{x^3}{x^4+4}\,dx &&\text{Let }&u=\unknown + \displaystyle \int \dfrac{x^3}{x^4+4}\,dx &&\text{Let }&u=\unknown -&&& du = \unknown\,dx + &&& du = \unknown\,dx -&&& \unknown \, du = \unknown\,dx + &&& \unknown \, du = \unknown\,dx -\displaystyle \int \dfrac{x^3}{x^4+4}\,dx &= \displaystyle \int \dfrac{\unknown}{\unknown}\, du + \displaystyle \int \dfrac{x^3}{x^4+4}\,dx &= \displaystyle \int \dfrac{\unknown}{\unknown}\, du -&= \cdots + &= \cdots -&= \dfrac{1}{4}\ln|x^4+4|+C + &= \dfrac{1}{4}\ln|x^4+4|+C     + + + + \displaystyle \int \dfrac{x^3}{x^4+4}\,dx &&\text{Let }&u=x^4 + 4 + + + &&& du = 4x^3 \,dx + + + &&& \dfrac{1}{4} \, du = x^3\,dx + + + \displaystyle \int \dfrac{x^3}{x^4+4}\,dx &= \displaystyle \int \dfrac{1/4}{u}\, du + + + &= \dfrac{1}{4}\ln|u| + C + + + &= \dfrac{1}{4}\ln|x^4+4|+C + + +     +

    -Given that -\displaystyle \int \dfrac{x^3}{x^4+4}\,dx -= \dfrac{1}{4}\ln|x^4+4|+C -, what is the value of -\displaystyle \int_0^2 \dfrac{x^3}{x^4+4}\,dx -? + Given that \displaystyle \int \dfrac{x^3}{x^4+4}\,dx = \dfrac{1}{4}\ln|x^4+4|+C , what is the value of \displaystyle \int_0^2 \dfrac{x^3}{x^4+4}\,dx?

    1. \dfrac{8}{20}

      2. @@ -381,6 +456,11 @@ Given that

    2. \dfrac{1}{4}\ln(4)-\dfrac{1}{4}\ln(20)

     + +

    + C: \dfrac{1}{4}\ln(20)-\dfrac{1}{4}\ln(4) +

    +     @@ -389,70 +469,74 @@ What's wrong with the following computation?

    -\displaystyle \int_0^2 \dfrac{x^3}{x^4+4}\,dx &&\text{Let }&u=x^4+4 + \displaystyle \int_0^2 \dfrac{x^3}{x^4+4}\,dx &&\text{Let }&u=x^4+4 -&&& du = 4x^3\,dx + &&& du = 4x^3\,dx -&&& \dfrac{1}{4} du = x^3\,dx + &&& \dfrac{1}{4} du = x^3\,dx -\displaystyle \int_0^2 \dfrac{x^3}{x^4+4}\,dx &= \displaystyle \int_0^2 \dfrac{1/4}{u}\, du + \displaystyle \int_0^2 \dfrac{x^3}{x^4+4}\,dx &= \displaystyle \int_0^2 \dfrac{1/4}{u}\, du -&= \left[\dfrac{1}{4}\ln|u|\right]_0^2 + &= \left[\dfrac{1}{4}\ln|u|\right]_0^2 -&= \dfrac{1}{4}\ln 2-\dfrac{1}{4}\ln 0 + &= \dfrac{1}{4}\ln 2-\dfrac{1}{4}\ln 0

    1. -The wrong u substitution was made. + The wrong u substitution was made.

    2. -The antiderivative of \dfrac{1/4}{u} was wrong. + The antiderivative of \dfrac{1/4}{u} was wrong.

    3. -The x values 0,2 were plugged in for the variable u. + The x values 0,2 were plugged in for the variable u.

    + +

    + C: The x values 0,2 were plugged in for the variable u. +

    +

    -Here's one way to show the computation of this definite integral by -tracking x values in the bounds. + Here's one way to show the computation of this definite integral by tracking x values in the bounds.

    -\displaystyle \int_0^2 \dfrac{x^3}{x^4+4}\,dx &&\text{Let }&u=x^4+4 + \displaystyle \int_0^2 \dfrac{x^3}{x^4+4}\,dx &&\text{Let }&u=x^4+4 -&&& du = 4x^3\,dx + &&& du = 4x^3\,dx -&&& \dfrac{1}{4} du = x^3\,dx + &&& \dfrac{1}{4} du = x^3\,dx -\displaystyle \int_{x=0}^{x=2} \dfrac{x^3}{x^4+4}\,dx &= \displaystyle \int_{x=0}^{x=2} \dfrac{1/4}{u}\, du + \displaystyle \int_{x=0}^{x=2} \dfrac{x^3}{x^4+4}\,dx &= \displaystyle \int_{x=0}^{x=2} \dfrac{1/4}{u}\, du -&= \left[ \dfrac{1}{4}\ln|u|\right]_{x=0}^{x=2} + &= \left[ \dfrac{1}{4}\ln|u|\right]_{x=0}^{x=2} -&= \left[ \dfrac{1}{4}\ln|x^4+4|\right]_{x=0}^{x=2} + &= \left[ \dfrac{1}{4}\ln|x^4+4|\right]_{x=0}^{x=2} -&= \dfrac{1}{4}\ln(20)-\dfrac{1}{4}\ln(4) + &= \dfrac{1}{4}\ln(20)-\dfrac{1}{4}\ln(4)     @@ -460,71 +544,92 @@ tracking x values in the bounds.

    -Instead of unsubstituting u values for x values, -definite integrals may be computed by also substituting x values in -the bounds with u values. Use this idea to complete -the following solution: + Instead of unsubstituting u values for x values, definite integrals may be computed by also substituting x values in the bounds with u values. Use this idea to complete the following solution:

    -\displaystyle \int_1^3 x^2e^{x^3}\,dx &&\text{Let }&u=\unknown + \displaystyle \int_1^3 x^2e^{x^3}\,dx &&\text{Let }&u=\unknown -&&&du = 3x^2\,dx + &&&du = 3x^2\,dx -&&&\dfrac{1}{3}du = x^2\,dx + &&&\dfrac{1}{3}du = x^2\,dx -\displaystyle \int_1^3 x^2e^{x^3}\,dx &= \displaystyle \int_{x=1}^{x=3} e^{(x^3)} (x^2\,dx) + \displaystyle \int_1^3 x^2e^{x^3}\,dx &= \displaystyle \int_{x=1}^{x=3} e^{(x^3)} (x^2\,dx) -&= \displaystyle \int_{u=\unknown}^{u=\unknown} e^{u} \dfrac{1}{3}\, du - --> + &= \displaystyle \int_{u=\unknown}^{u=\unknown} e^{u} \dfrac{1}{3}\, du + -&= \left[\dfrac{1}{3}e^{u}\right]_{\unknown}^{\unknown} + &= \left[\dfrac{1}{3}e^{u}\right]_{\unknown}^{\unknown} -&= \unknown + &= \unknown     + +

    + + + \displaystyle \int_1^3 x^2e^{x^3}\,dx &&\text{Let }&u=x^3 + + + &&&du = 3x^2\,dx + + + &&&\dfrac{1}{3}du = x^2\,dx + + + \displaystyle \int_1^3 x^2e^{x^3}\,dx &= \displaystyle \int_{x=1}^{x=3} e^{(x^3)} (x^2\,dx) + + + &= \displaystyle \int_{u=1}^{u=27} e^{u} \dfrac{1}{3}\, du + + + &= \left[\dfrac{1}{3}e^{u}\right]_{1}^{27} + + + &= \dfrac{1}{3}e^{27} - \dfrac{1}{3}e + + +

    +

    -Here is how one might write out the explanation -of how to find -\displaystyle \int_1^3 x^2e^{x^3}\,dx from start to finish by -leaving bounds in terms of x instead: + Here is how one might write out the explanation of how to find \displaystyle \int_1^3 x^2e^{x^3}\,dx from start to finish by leaving bounds in terms of x instead:

    -\displaystyle \int_1^3 x^2e^{x^3}\,dx &&\text{Let }&u=x^3 + \displaystyle \int_1^3 x^2e^{x^3}\,dx &&\text{Let }&u=x^3 -&&& du = 3x^2\,dx + &&& du = 3x^2\,dx -&&& \dfrac{1}{3}du = x^2\,dx + &&& \dfrac{1}{3}du = x^2\,dx -\displaystyle \int_1^3 x^2e^{x^3}\,dx &= \displaystyle \int_{x=1}^{x=3} e^{(x^3)} (x^2\,dx) + \displaystyle \int_1^3 x^2e^{x^3}\,dx &= \displaystyle \int_{x=1}^{x=3} e^{(x^3)} (x^2\,dx) -&= \displaystyle \int_{x=1}^{x=3} e^{u} \dfrac{1}{3}\, du + &= \displaystyle \int_{x=1}^{x=3} e^{u} \dfrac{1}{3}\, du -&= \left[\dfrac{1}{3}e^{u}\right]_{x=1}^{x=3} + &= \left[\dfrac{1}{3}e^{u}\right]_{x=1}^{x=3} -&= \left[\dfrac{1}{3}e^{x^3}\right]_{x=1}^{x=3} + &= \left[\dfrac{1}{3}e^{x^3}\right]_{x=1}^{x=3} -&= \dfrac{1}{3}e^{3^3} - \dfrac{1}{3}e^{1^3} + &= \dfrac{1}{3}e^{3^3} - \dfrac{1}{3}e^{1^3} -&= \dfrac{1}{3}e^{27} - \dfrac{1}{3}e + &= \dfrac{1}{3}e^{27} - \dfrac{1}{3}e     @@ -532,71 +637,101 @@ leaving bounds in terms of x instead:

    -Use substitution to show that -\displaystyle \int_1^4 \dfrac{e^{\sqrt x}}{\sqrt x}\,dx=2e^2-2e. + Use substitution to show that + \displaystyle \int_1^4 \dfrac{e^{\sqrt x}}{\sqrt x}\,dx=2e^2-2e.

    -Use substitution to show that -\displaystyle \int_0^{\pi/4} \sin(2\theta)\,d\theta=\dfrac{1}{2}. + Use substitution to show that + \displaystyle \int_0^{\pi/4} \sin(2\theta)\,d\theta=\dfrac{1}{2}.

    +

    -Use substitution to show that + Use substitution to show that -\displaystyle \int u^5(u^3+1)^{1/3}\,du= -\dfrac{1}{7}(u^3+1)^{7/3}- -\dfrac{1}{4}(u^3+1)^{4/3}+C + \displaystyle \int u^5(u^3+1)^{1/3}\,du= \dfrac{1}{7}(u^3+1)^{7/3} - \dfrac{1}{4}(u^3+1)^{4/3}+C .

    +

    -Consider \displaystyle \int (3x-5)^2\,dx. + Consider \displaystyle \int (3x-5)^2\,dx.

    -

    -Solve this integral using substitution. -

    + +

    + Solve this integral using substitution. +

    +     + +

    + \displaystyle \int (3x-5)^2\, dx = \dfrac{1}{9}(3x-5)^3 + C +

    +     -

    -Replace (3x-5)^2 with (9x^2-30x+25) -in the original integral, the solve using -the reverse power rule. -

    + +

    + Replace (3x-5)^2 with (9x^2-30x+25) in the original integral, the solve using the reverse power rule. +

    +     + +

    + \displaystyle \int 9x^2 - 30x + 25\, dx = 3x^3 - 15x^2 + 25x + C +

    +     + +

    + Which method did you prefer? +

    +

    -Which method did you prefer? +

    -Consider \displaystyle \int \tan(x)\,dx. + Consider \displaystyle \int \tan(x)\,dx.

    -

    -Replace \tan(x) in the integral with -a fraction involving sine and cosine. -

    + +

    + Replace \tan(x) in the integral with a fraction involving sine and cosine. +

    +     + +

    + \displaystyle \int \tan(x)\, dx = \int \dfrac{\sin(x)}{\cos(x)}\, dx +

    +     -

    -Use substitution to solve the integral. -

    + +

    + Use substitution to solve the integral. +

    +     + +

    + \displaystyle \int \dfrac{\sin(x)}{\cos(x)}\, dx = -\ln |\cos(x)| + C or \ln |\sec(x)| + C +

    +     From 165d1a84505e5eea353e417b64a43fb9e2b5e2fc Mon Sep 17 00:00:00 2001 From: Cory Wilson <40644848+cg2wilson@users.noreply.github.com> Date: Thu, 22 May 2025 19:09:41 +0000 Subject: [PATCH 2/4] add answers for TI2 --- source/calculus/source/05-TI/02.ptx | 432 ++++++++++++++++++++++------ 1 file changed, 348 insertions(+), 84 deletions(-) diff --git a/source/calculus/source/05-TI/02.ptx b/source/calculus/source/05-TI/02.ptx index 99e3c4de1..086fe133c 100644 --- a/source/calculus/source/05-TI/02.ptx +++ b/source/calculus/source/05-TI/02.ptx @@ -11,63 +11,121 @@ Activities - + +

    Answer the following.

     + -

    Using the product rule, which of these is derivative of x^3e^x with respect to x?

    -
      -

    1. 3x^2e^x

      2. -

    2. 3x^2e^{x}+x^3e^x

      3. -

    3. 3x^2e^{x-1}

      4. -

    4. \dfrac{1}{4}x^4 e^x

      5. -
    + +

    + Using the product rule, which of these is derivative of x^3e^x with respect to x? +

      +

    1. 3x^2e^x

      2. +

    2. 3x^2e^{x}+x^3e^x

      3. +

    3. 3x^2e^{x-1}

      4. +

    4. \dfrac{1}{4}x^4 e^x

      5. +
    +

    +     + +

    + B: 3x^2 e^x + x^3e^x +

    +     -

    Based on this result, which of these would you suspect to equal \displaystyle \int 3x^2e^x+x^3e^x\,dx?

    -
      -

    1. x^3e^x+C

      2. -

    2. x^3e^x+\dfrac{1}{4}x^4e^x+C

      3. -

    3. 6xe^x+3x^2e^x+C

      4. -

    4. 6xe^x+3x^2e^x+3x^2e^x+x^3e^x+C

      5. -
    + +

    + Based on this result, which of these would you suspect to equal \displaystyle \int 3x^2e^x+x^3e^x\,dx? +

      +

    1. x^3e^x+C

      2. +

    2. x^3e^x+\dfrac{1}{4}x^4e^x+C

      3. +

    3. 6xe^x+3x^2e^x+C

      4. +

    4. 6xe^x+3x^2e^x+3x^2e^x+x^3e^x+C

      5. +
    +

    +     + +

    + A: x^3e^x + C +

    +     +

    Answer the following.

     -

    Which differentiation rule is easier to implement?

    -
      -

    1. Product Rule

      2. -

    2. Chain Rule

      3. -
    + +

    + Which differentiation rule is easier to implement? +

      +

    1. Product Rule

      2. +

    2. Chain Rule

      3. +
    +

    +     + +

    + Answers vary, but probably A: Product Rule. +

    +     -

    Which differentiation strategy do you expect to be easier to reverse?

    -
      -

    1. Product Rule

      2. -

    2. Chain Rule

      3. -
    + +

    + Which differentiation strategy do you expect to be easier to reverse? +

      +

    1. Product Rule

      2. +

    2. Chain Rule

      3. +
    +

    +     + +

    + Answers vary, but likely A: Product Rule. +

    +     +

    Answer the following.

     -

    Which of the following equations is equivalent to the formula \dfrac{d}{dx}[uv]=u'v+uv'?

    -
      -

    1. uv'=-\dfrac{d}{dx}(uv)-vu'

      2. -

    2. uv'=-\dfrac{d}{dx}(uv)+vu'

      3. -

    3. uv'=\dfrac{d}{dx}(uv)+vu'

      4. -

    4. uv'=\dfrac{d}{dx}(uv)-vu'

      5. -
    + +

    + Which of the following equations is equivalent to the formula \dfrac{d}{dx}[uv]=u'v+uv'? +

      +

    1. uv'=-\dfrac{d}{dx}(uv)-vu'

      2. +

    2. uv'=-\dfrac{d}{dx}(uv)+vu'

      3. +

    3. uv'=\dfrac{d}{dx}(uv)+vu'

      4. +

    4. uv'=\dfrac{d}{dx}(uv)-vu'

      5. +
    +

    +     + +

    + D: uv'=\dfrac{d}{dx}(uv)-vu' +

    +     -

    Which of these is the most concise result of integrating both sides with respect to x?

    -
      -

    1. \displaystyle \int(uv')\,dx=uv-\int(vu')\,dx

      2. -

    2. \displaystyle \int(u)\,dv=uv-\int(v)\,du

      3. -

    3. \displaystyle \int(uv')\,dx=uv-\int(vu')\,dx+C

      4. -

    4. \displaystyle \int(u)\,dv=uv-\int(v)\,du+C

      5. -
    + +

    + Which of these is the most concise result of integrating both sides with respect to x? +

      +

    1. \displaystyle \int(uv')\,dx=uv-\int(vu')\,dx

      2. +

    2. \displaystyle \int(u)\,dv=uv-\int(v)\,du

      3. +

    3. \displaystyle \int(uv')\,dx=uv-\int(vu')\,dx+C

      4. +

    4. \displaystyle \int(u)\,dv=uv-\int(v)\,du+C

      5. +
    +

    +     + +

    + B: \displaystyle \int(u)\,dv=uv-\int(v)\,du +

    +     @@ -84,51 +142,100 @@

    Consider \displaystyle \int xe^{x}\,dx. Suppose we decided to let u=x.

    -

    Compute \dfrac{du}{dx}=\unknown, and rewrite it as du=\unknown\,dx.

    + +

    + Compute \dfrac{du}{dx}=\unknown, and rewrite it as du=\unknown\,dx. +

    +     + +

    + \dfrac{du}{dx} = 1, so du = 1\, dx. +

    +     -

    What is the best candidate for dv?

    -
      -

    1. dv=x\,dx

      2. -

    2. dv=e^x

      3. -

    3. dv=x

      4. -

    4. dv=e^x\,dx

      5. -
    + +

    + What is the best candidate for dv? +

      +

    1. dv=x\,dx

      2. +

    2. dv=e^x

      3. +

    3. dv=x

      4. +

    4. dv=e^x\,dx

      5. +
    +

    +     + +

    + D: dv=e^x\,dx +

    +     -

    Given that dv=e^x\,dx, find v=\unknown.

    + +

    + Given that dv=e^x\,dx, find v=\unknown. +

    +     + +

    + v = e^x +

    +     -

    Show why \displaystyle \int xe^{x}\,dx may now be rewritten as \displaystyle xe^x-\int e^x\,dx.

    + +

    + Show why \displaystyle \int xe^{x}\,dx may now be rewritten as \displaystyle xe^x-\int e^x\,dx. +

    +     + +

    + Since u = x, du = dx, dv = e^x\, dx, and v = e^x, the formula for integration by parts gives us + \int xe^x\, dx = xe^x-\int e^x\, dx +

    +     -

    Solve \displaystyle \int e^x\,dx, and then give the most general antiderivative of \displaystyle \int xe^{x}\,dx.

    + +

    + Solve \displaystyle \int e^x\,dx, and then give the most general antiderivative of \displaystyle \int xe^{x}\,dx. +

    +     + +

    + \displaystyle \int e^x\, dx = e^x so that \displaystyle \int xe^x\, dx = xe^x - e^x + C +

    +     + -

    Here is how one might write out the explanation of how to find \displaystyle \int xe^{x}\,dx from start to finish: +

    + Here is how one might write out the explanation of how to find \displaystyle \int xe^{x}\,dx from start to finish:

    -\displaystyle \int xe^{x}\,dx + \displaystyle \int xe^{x}\,dx -& u=x & dv = e^x\,dx + & u=x & dv = e^x\,dx -& du=1\cdot\,dx & v=e^x + & du=1\cdot\,dx & v=e^x -\displaystyle \int xe^x \,dx &= xe^x-\int e^x \,dx + \displaystyle \int xe^x \,dx &= xe^x-\int e^x \,dx -&= xe^{x}-e^x+C + &= xe^{x}-e^x+C     +

    Which step of the previous example do you think was the most important?

    @@ -148,29 +255,74 @@     + +

    + A: Choosing u=x and dv = e^x\, dx +

    +     + -

    Consider the integral \displaystyle \int x^9\ln(x) \,dx. Suppose we proceed using integration by parts. We choose u=\ln(x) and dv=x^9\,dx.

    +

    + Consider the integral \displaystyle \int x^9\ln(x) \,dx. Suppose we proceed using integration by parts. We choose u=\ln(x) and dv=x^9\,dx. +

    -

    What is du?

    + +

    + What is du? +

    +     + +

    + +

    +     >     -

    What is v?

    + +

    + What is v? +

    +     + +

    + +

    +     >     -

    What do you get when plugging these pieces into integration by parts?

    + +

    + What do you get when plugging these pieces into integration by parts? +

    +     + +

    + +

    +     >     -

    Does the new integral \displaystyle \int v\,du seem easier or harder to compute than the original integral \displaystyle \int x^9\ln(x) \,dx?

    -
      -

    1. The original integral is easier to compute.

      2. -

    2. The new integral is easier to compute.

      3. -

    3. Neither integral seems harder than the other one.

      4. -
    + +

    + Does the new integral \displaystyle \int v\,du seem easier or harder to compute than the original integral \displaystyle \int x^9\ln(x) \,dx? +

      +

    1. The original integral is easier to compute.

      2. +

    2. The new integral is easier to compute.

      3. +

    3. Neither integral seems harder than the other one.

      4. +
    +

    +     + +

    + B: The new integral is easier to compute. +

    +     +

    Consider the integral \displaystyle \int x^9\ln(x) \,dx once more. Suppose we still proceed using integration by parts. However, this time we choose u=x^9 and dv=\ln(x)\,dx. Do you prefer this choice or the choice we made in ?

    @@ -180,10 +332,16 @@

  • We do not have a strong preference, since these choices are of the same difficulty.

    •  + +

    + A: We prefer the substitution choice of u=\ln(x) and dv=x^9\,dx. +

    +     + -

    Consider the integral \int x\cos(x)\,dx. Suppose we proceed using integration by parts. Which of the following candidates for u and dv would best allow you to evaluate this integral?

    +

    Consider the integral \displaystyle \int x\cos(x)\,dx. Suppose we proceed using integration by parts. Which of the following candidates for u and dv would best allow you to evaluate this integral?

    1. u=\cos(x), dv=x\, dx

      @@ -199,17 +357,35 @@
    + +

    + D: u=x, dv=\cos(x)\,dx +

    +     +

    Evaluate the integral \displaystyle \int x\cos(x)\,dx using integration by parts.

     + +

    + \displaystyle \int x\cos(x)\, dx = x\sin x + \cos(x) + C +

    +     +

    Now use integration by parts to evaluate the integral \displaystyle \int_{\pi/6}^{\pi} x\cos(x)\,dx.

     + +

    + \displaystyle \int_{\pi/6}^{\pi} x\cos(x)\,dx = -\dfrac{\pi}{12} - \sqrt{3}{2} +

    +     +

    Consider the integral \displaystyle \int x\arctan(x)\,dx. Suppose we proceed using integration by parts. Which of the following candidates for u and dv would best allow you to evaluate this integral?

    @@ -228,7 +404,13 @@     + +

    + B: u=\arctan(x), dv=x\,dx +

    +     +

    Consider the integral \displaystyle \int e^x\cos(x)\,dx. Suppose we proceed using integration by parts. Which of the following candidates for u and dv would best allow you to evaluate this integral?

    @@ -247,25 +429,30 @@     + +

    + Answers vary +

    +

    Suppose we started using integration by parts to solve the integral \displaystyle \int e^x\cos(x)\,dx as follows:

    -\int e^x\cos(x)\,dx + \int e^x\cos(x)\,dx -& u=\cos(x) & dv = e^x\,dx + & u=\cos(x) & dv = e^x\,dx -& du=-\sin(x) \,dx & v=e^x + & du=-\sin(x) \,dx & v=e^x -\int e^x\cos(x)\,dx &= \cos(x)e^x-\int e^x(-\sin(x) \,dx) + \int e^x\cos(x)\,dx &= \cos(x)e^x-\int e^x(-\sin(x) \,dx) -&= \cos(x)e^x+\int e^x\sin(x) \,dx + &= \cos(x)e^x+\int e^x\sin(x) \,dx

    We will have to use integration by parts a second time to evaluate the integral \displaystyle \int e^x\sin(x) \,dx. Which of the following candidates for u and dv would best allow you to continue evaluating the original integral \displaystyle \int e^x\cos(x)\,dx?

    @@ -284,35 +471,86 @@     + +

    + B: u=\sin(x), dv=e^x\,dx +

    +     +

    Use integration by parts to show that \displaystyle \int_0^{\pi/4} x\sin(2x)\,dx=\dfrac{1}{4}.

     + +

    + Use u = x and dv = \sin (2x)\, dx to get the result. +

    +     +

    Consider the integral \displaystyle \int t^5 \sin(t^3)\,dt.

     -

    Use the substitution x=t^3 to rewrite the integral in terms of x.

    + +

    + Use the substitution x=t^3 to rewrite the integral in terms of x. +

    +     + +

    + \displaystyle \int t^5\sin(t^3)\, dt = \int \dfrac{1}{3}x\sin(x)\, dx +

    +     >     -

    Use integration by parts to evaluate the integral in terms of x.

    + +

    + Use integration by parts to evaluate the integral in terms of x. +

    +     + +

    + \displaystyle \int \dfrac{1}{3}x\sin(x) = -\dfrac{1}{3}x\cos(x) + \dfrac{1}{3}\sin(x) + C +

    +     >     -

    Replace x with t^3 to finish evaluating the original integral.

    + +

    + Replace x with t^3 to finish evaluating the original integral. +

    +     + +

    + \displaystyle \int t^5\sin(t^3)\, dt = -\dfrac{1}{3}t^3\cos(t^3) + \dfrac{1}{3}\sin(t^3) + C +

    +     >     +

    Use integration by parts to show that \displaystyle \int \ln(z)\,dz=z \ln(z) - z + C.

     + +

    + Use u = \ln z and dv = dz to get the result. +

    +     +

    Given that that \displaystyle \int \ln(z)\,dz=z \ln(z) - z + C, evaluate \displaystyle \int (\ln(z))^2\,dz.

     + +

    + \displaystyle \int (\ln(x))^2\, dz = z(\ln(z))^2 - 2z\ln(z) + 2z + C +

    +     @@ -323,27 +561,48 @@

    -

    - Noting that \displaystyle\int (\sin(x))^2\, dx=\int (\sin(x))(\sin(x))dx and letting u=\sin(x), dv=\sin(x)\, dx, what equality does integration by parts yield? + +

    + Noting that \displaystyle\int (\sin(x))^2\, dx=\int (\sin(x))(\sin(x))dx and letting u=\sin(x), dv=\sin(x)\, dx, what equality does integration by parts yield?

    1. \displaystyle\int (\sin(x))^2dx=\sin(x)\cos(x)+\int (\cos(x))^2\, dx.
    2. \displaystyle\int (\sin(x))^2dx=-\sin(x)\cos(x)+\int (\cos(x))^2\, dx.
    3. \displaystyle\int (\sin(x))^2dx=\sin(x)\cos(x)-\int (\cos(x))^2\, dx.
    4. \displaystyle\int (\sin(x))^2dx=-\sin(x)\cos(x)-\int (\cos(x))^2\, dx.
    -

    +

    +     + +

    + B: \displaystyle\int (\sin(x))^2dx=-\sin(x)\cos(x)+\int (\cos(x))^2\, dx. +

    +     -

    - Use the fact that (\cos(x))^2=1-(\sin(x))^2 to rewrite the above equality. -

    + +

    + Use the fact that (\cos(x))^2=1-(\sin(x))^2 to rewrite the above equality. +

    +     + +

    + \displaystyle\int (\sin(x))^2dx=-\sin(x)\cos(x)+\int 1-(\sin(x))^2\, dx. +

    +     -

    - Solve algebraically for \displaystyle\int (\sin(x))^2\, dx. -

    + +

    + Solve algebraically for \displaystyle\int (\sin(x))^2\, dx. +

    +     + +

    + \displaystyle \int (\sin(x))^2\, dx = \dfrac{1}{2}x - \dfrac{1}{2}\sin(x)\cos(x) +

    +     @@ -356,6 +615,11 @@ Modifying the approach from , use parts to find \displaystyle\int (\cos(x))^2\, dx.

    + +

    + \displaystyle \int (\sin(x))^2\, dx = \dfrac{1}{2}x + \dfrac{1}{2}\sin(x)\cos(x) +

    +     From 0631a210c9ebcf6b917ed9b8a1206bd487da21d8 Mon Sep 17 00:00:00 2001 From: Cory Wilson <40644848+cg2wilson@users.noreply.github.com> Date: Thu, 22 May 2025 20:14:54 +0000 Subject: [PATCH 3/4] add answers for TI3 --- source/calculus/source/05-TI/03.ptx | 149 +++++++++++++++++++++++----- 1 file changed, 122 insertions(+), 27 deletions(-) diff --git a/source/calculus/source/05-TI/03.ptx b/source/calculus/source/05-TI/03.ptx index 6dc467f2a..99206db27 100644 --- a/source/calculus/source/05-TI/03.ptx +++ b/source/calculus/source/05-TI/03.ptx @@ -24,6 +24,11 @@

    + +

    + A or B +

    +     @@ -38,6 +43,11 @@

    + +

    + A: u = \sin(x) +

    +     @@ -52,76 +62,106 @@

    + +

    + D: Substitution is not effective +

    +

    - It's possible to use substitution to evaluate \displaystyle\int \sin^4(x)\cos^3(x) \, dx, - by taking advantage of the trigonometric identity \sin^2(x)+\cos^2(x)=1. + It's possible to use substitution to evaluate \displaystyle\int \sin^4(x)\cos^3(x) \, dx, by taking advantage of the trigonometric identity \sin^2(x)+\cos^2(x)=1.

    - Complete the following substitution of u=\sin(x),\, du=\cos(x)\,dx - by filling in the missing \unknowns. + Complete the following substitution of u=\sin(x),\, du=\cos(x)\,dx by filling in the missing \unknowns.

    -\int \sin^4(x)\cos^3(x)\,dx &=\int\sin^4(x)(\,\unknown\,)\cos(x)\,dx + \int \sin^4(x)\cos^3(x)\,dx &=\int\sin^4(x)(\,\unknown\,)\cos(x)\,dx -&=\int\sin^4(x)(1-\unknown)\cos(x)\,dx + &=\int\sin^4(x)(1-\unknown)\cos(x)\,dx -&= \int\unknown(1-\unknown)\,du + &= \int\unknown(1-\unknown)\,du -&= \int (u^4-u^6)\,du + &= \int (u^4-u^6)\,du -&= \frac{1}{5}u^5-\frac{1}{7}u^7+C + &= \frac{1}{5}u^5-\frac{1}{7}u^7+C -&= \unknown + &= \unknown     + + + + \int \sin^4(x)\cos^3(x)\,dx &=\int\sin^4(x)(\cos^2(x))\cos(x)\,dx + + + &=\int\sin^4(x)(1-\sin^2(x))\cos(x)\,dx + + + &= \int u^4(1-u^2)\,du + + + &= \int (u^4-u^6)\,du + + + &= \frac{1}{5}u^5-\frac{1}{7}u^7+C + + + &= \dfrac{1}{5}\sin^5(x) - \dfrac{1}{7}\sin^7(x) + C + + +

    -Trying to substitute u=\cos(x),du=-\sin(x)\,dx in the previous example is less successful. + Trying to substitute u=\cos(x),du=-\sin(x)\,dx in the previous example is less successful.

    -\int \sin^4(x)\cos^3(x)\,dx &=-\int\sin^3(x)\cos^3(x)(-\sin(x)\,dx) + \int \sin^4(x)\cos^3(x)\,dx &=-\int\sin^3(x)\cos^3(x)(-\sin(x)\,dx) -&=-\int\sin^3(x)u^3\,du + &=-\int\sin^3(x)u^3\,du -&= \cdots? + &= \cdots?

    -Which feature of \sin^4(x)\cos^3(x) made u=\sin(x) the better choice? + Which feature of \sin^4(x)\cos^3(x) made u=\sin(x) the better choice?

    1. The even power of \sin^4(x)
    2. The odd power of \cos^3(x)

     + +

    + B: The odd power of \cos^3(x) +

    +

    -Try to show \int \sin^5(x)\cos^2(x)\,dx= --\frac{1}{7} \, \cos^{7}\left(x\right) + \frac{2}{5} \, \cos^{5}\left(x\right) - \frac{1}{3} \, \cos^{3}\left(x\right)+C -by first trying u=\sin(x), and then trying u=\cos(x) instead. + Try to show + \int \sin^5(x)\cos^2(x)\,dx= -\frac{1}{7} \, \cos^{7}\left(x\right) + \frac{2}{5} \, \cos^{5}\left(x\right) - \frac{1}{3} \, \cos^{3}\left(x\right)+C + by first trying u=\sin(x), and then trying u=\cos(x) instead.

    -Which substitution worked better and why? + Which substitution worked better and why?

    1. u=\sin(x) due to \sin^5(x)'s odd power.
    2. u=\sin(x) due to \cos^2(x)'s even power.
      3. @@ -130,24 +170,27 @@ Which substitution worked better and why?

    + +

    + C: u=\cos(x) due to \sin^5(x)'s odd power. +

    +

    -When integrating the form \displaystyle \int \sin^m(x)\cos^n(x)\,dx: + When integrating the form \displaystyle \int \sin^m(x)\cos^n(x)\,dx:

    • -If \sin's power is odd, rewrite the integral as -\displaystyle \int g(\cos(x))\sin(x)\,dx and use u=\cos(x). + If \sin's power is odd, rewrite the integral as \displaystyle \int g(\cos(x))\sin(x)\,dx and use u=\cos(x).

    • -If \cos's power is odd, rewrite the integral as -\displaystyle \int h(\sin(x))\cos(x)\,dx and use u=\sin(x). + If \cos's power is odd, rewrite the integral as \displaystyle \int h(\sin(x))\cos(x)\,dx and use u=\sin(x).

    @@ -163,6 +206,11 @@ If \cos's power is odd, rewrite the integral as

    Use the fact that \sin^2(\theta)=\displaystyle\frac{1-\cos(2\theta)}{2} to rewrite the integrand using the above identities as an integral involving \cos(2x).

     + +

    + \displaystyle \int \sin^2(x)\, dx = \int \dfrac{1}{2}(1-\cos (2x))\, dx +

    +     @@ -179,11 +227,21 @@ If \cos's power is odd, rewrite the integral as

    Use the fact that \cos^2(\theta)=\displaystyle\frac{1+\cos(2\theta)}{2} and \sin^2(\theta)=\displaystyle\frac{1-\cos(2\theta)}{2} to rewrite the integrand using the above identities as an integral involving \cos^2(2x).

     + +

    + \displaystyle \int \sin^2(x)\cos^2(x)\, dx = \int \dfrac{1}{4}(1-\cos(2x))(1+\cos(2x))\, dx +

    +

    Use the above identities to rewrite this new integrand as one involving \cos(4x).

     + +

    + \displaystyle \int \dfrac{1}{4}(1-\cos(2x))(1+\cos(2x))\, dx = \dfrac{1}{4}\int 1-\cos^2(2x)\, dx = \dfrac{1}{4}\int 1-\dfrac{1}{2}(1+\cos(4x))\, dx +

    +     @@ -204,6 +262,11 @@ If \cos's power is odd, rewrite the integral as

    + +

    + C: \displaystyle\int \left(\frac{1-\cos(2x)}{2}\right)^2\left(\frac{1+\cos(2x)}{2}\right)^2 \, dx +

    +     @@ -218,6 +281,11 @@ If \cos's power is odd, rewrite the integral as

    + +

    + A or C +

    +     @@ -239,13 +307,40 @@ We might also use some other trigonometric identities to manipulate our integran

    -

    Find an identity from which could be used to transform our integrand.

    + +

    + Find an identity from which could be used to transform our integrand. +

    +     + +

    + \sin(\alpha)\sin(\beta) = \dfrac{\cos(\alpha-\beta) - \cos(\alpha + \beta)}{2} +

    +     >     -

    Rewrite the integrand using the selected identity.

    + +

    + Rewrite the integrand using the selected identity. +

    +     + +

    + \sin(\theta)\sin(3\theta) = \dfrac{1}{2}(\cos(\theta - 3\theta) - \cos(\theta + 3\theta) = \dfrac{1}{2}(\cos (-2\theta) - \cos(4\theta)) +

    +     -

    Evaluate the integral.

    + +

    + Evaluate the integral. +

    +     + +

    + \displaystyle \int \sin(\theta)\sin(3\theta)\, d\theta = \int \dfrac{1}{2}(\cos (-2\theta) - \cos(4\theta)) = \dfrac{1}{4}\sin(2\theta) - \dfrac{1}{8}\sin(4\theta) + C +

    +     >     From f2abf9b5e23b25fe485cb9cfab634910d8582575 Mon Sep 17 00:00:00 2001 From: Cory Wilson <40644848+cg2wilson@users.noreply.github.com> Date: Thu, 22 May 2025 22:13:08 +0000 Subject: [PATCH 4/4] add answers to TI4 --- source/calculus/source/05-TI/04.ptx | 341 ++++++++++++++++++++++------ 1 file changed, 271 insertions(+), 70 deletions(-) diff --git a/source/calculus/source/05-TI/04.ptx b/source/calculus/source/05-TI/04.ptx index ca5bbe2e3..c196626ed 100644 --- a/source/calculus/source/05-TI/04.ptx +++ b/source/calculus/source/05-TI/04.ptx @@ -23,6 +23,11 @@

    + +

    + D: Substitution is not effective. +

    +     @@ -38,6 +43,11 @@

    + +

    + B: Let \theta satisfy 9-4x^2=9-9\sin^2\theta=9\cos^2\theta. +

    +     @@ -46,28 +56,54 @@ Fill in the missing \unknowns for the following calculation. -\text{Let }9-4x^2&=9-9\sin^2\theta=9\cos^2\theta + \text{Let }9-4x^2&=9-9\sin^2\theta=9\cos^2\theta -4x^2&=\unknown + 4x^2&=\unknown -x&=\unknown + x&=\unknown -dx&=\unknown\,d\theta + dx&=\unknown\,d\theta -\int\sqrt{9-4x^2} \,dx&=\int\sqrt{\unknown}\,(\unknown\,d\theta) + \int\sqrt{9-4x^2} \,dx&=\int\sqrt{\unknown}\,(\unknown\,d\theta) -&= \int\frac{9}{2}\cos^2 \theta\,d\theta + &= \int\frac{9}{2}\cos^2 \theta\,d\theta

    + +

    + + + \text{Let }9-4x^2&=9-9\sin^2\theta=9\cos^2\theta + + + 4x^2&=9\sin^2\theta + + + x&=\dfrac{3}{2}\sin\theta + + + dx&=\dfrac{3}{2}\cos\theta\,d\theta + + + + + \int\sqrt{9-4x^2} \,dx&=\int\sqrt{9-9\sin^2\theta}\,\left((\dfrac{3}{2}\cos\theta\,d\theta\right) + + + &= \int\frac{9}{2}\cos^2 \theta\,d\theta + + +

    +     @@ -80,27 +116,51 @@ dx&=\unknown\,d\theta

    -\sin(\theta)&=\unknown + \sin(\theta)&=\unknown -\theta&=\arcsin(\unknown) + \theta&=\arcsin(\unknown) -\cos(\theta)&=\unknown\sqrt{\unknown} + \cos(\theta)&=\unknown\sqrt{\unknown} -\int\sqrt{9-4x^2} \,dx&= \cdots = \int\frac{9}{2}\cos^2 \theta\,d\theta + \int\sqrt{9-4x^2} \,dx&= \cdots = \int\frac{9}{2}\cos^2 \theta\,d\theta -&= \frac{9}{2}\left(\frac{1}{2}\theta+\frac{1}{2}\sin\theta\cos\theta\right)+C + &= \frac{9}{2}\left(\frac{1}{2}\theta+\frac{1}{2}\sin\theta\cos\theta\right)+C -&= \frac{9}{4}(\unknown)+\dfrac{9}{4}(\unknown)(\unknown)+C + &= \frac{9}{4}(\unknown)+\dfrac{9}{4}(\unknown)(\unknown)+C + + + + \sin(\theta)&= \dfrac{2}{3}x + + + \theta&=\arcsin\left(\dfrac{2}{3}x\right) + + + \cos(\theta)&= \dfrac{1}{3}\sqrt{9-4x^2} + + + + + \int\sqrt{9-4x^2} \,dx&= \cdots = \int\frac{9}{2}\cos^2 \theta\,d\theta + + + &= \frac{9}{2}\left(\frac{1}{2}\theta+\frac{1}{2}\sin\theta\cos\theta\right)+C + + + &= \frac{9}{4}\arcsin\left(\dfrac{2}{3}x\right)+\dfrac{9}{4}\left(\dfrac{2}{3}x\right)\left(\dfrac{1}{3}\sqrt{9-4x^2}\right)+C + + +     @@ -109,48 +169,81 @@ dx&=\unknown\,d\theta Use similar reasoning to complete the following proof that \dfrac{d}{dx}\left[\arcsin(x)\right]=\dfrac{1}{\sqrt{1-x^2}}. -\text{Let }1-x^2&=1-\unknown\theta=\unknown\theta + \text{Let }1-x^2&=1-\unknown\theta=\unknown\theta -x^2&=\unknown + x^2&=\unknown -x&=\unknown + x&=\unknown -dx&=\unknown\,d\theta + dx&=\unknown\,d\theta -\theta&=\unknown + \theta&=\unknown -\int \dfrac{1}{\sqrt{1-x^2}} \,dx&=\displaystyle \int\frac{1}{\sqrt{\unknown}}\,(\unknown\,d\theta) + \int \dfrac{1}{\sqrt{1-x^2}} \,dx&=\displaystyle \int\frac{1}{\sqrt{\unknown}}\,(\unknown\,d\theta) -&= \int \,d\theta + &= \int \,d\theta -&= \unknown + C + &= \unknown + C -&= \arcsin(x) + C + &= \arcsin(x) + C

    + +

    + + + \text{Let }1-x^2&=1-\sin^2\theta=\cos^2\theta + + + x^2&=\sin^2\theta + + + x&=\sin\theta + + + dx&=\cos\theta\,d\theta + + + \theta&=\arcsin(x) + + + + + \int \dfrac{1}{\sqrt{1-x^2}} \,dx&=\displaystyle \int\frac{1}{\sqrt{\cos^2\theta}}\,(\cos\theta\,d\theta) + + + &= \int \,d\theta + + + &= \theta + C + + + &= \arcsin(x) + C + + +

    +

    -Substitutions of the form 16-25x^2=16-16\sin^2x=16\cos^2x are made possible due to the -Pythagorean identity \sin^2(x)+\cos^2(x)=1. + Substitutions of the form 16-25x^2=16-16\sin^2x=16\cos^2x are made possible due to the Pythagorean identity \sin^2(x)+\cos^2(x)=1.

    -Which two of these four identities can be obtained from dividing both sides of -\sin^2(x)+\cos^2(x)=1 by \cos^2(x) and rearranging? + Which two of these four identities can be obtained from dividing both sides of \sin^2(x)+\cos^2(x)=1 by \cos^2(x) and rearranging?

    1. \tan^2(x)-1=\sec^2(x)
    2. \tan^2(x)+1=\sec^2(x)
      3. @@ -159,35 +252,39 @@ Which two of these four identities can be obtained from dividing both sides of

    + +

    + B: \tan^2(x) + 1 = \sec^2(x) +

    +

    -In summary, certain quadratic expressions inside an integral may be substituted with -trigonometric functions to take advantage of trigonometric identities and simplify the integrand: + In summary, certain quadratic expressions inside an integral may be substituted with trigonometric functions to take advantage of trigonometric identities and simplify the integrand: -\text{Let } b^2-a^2x^2&=b^2-b^2\sin^2(\theta)=b^2\cos^2(\theta) + \text{Let } b^2-a^2x^2&=b^2-b^2\sin^2(\theta)=b^2\cos^2(\theta) -\text{So } x&=\frac{b}{a}\sin(\theta) + \text{So } x&=\frac{b}{a}\sin(\theta) -\text{Let } b^2+a^2x^2&=b^2+b^2\tan^2(\theta)=b^2\sec^2(\theta) + \text{Let } b^2+a^2x^2&=b^2+b^2\tan^2(\theta)=b^2\sec^2(\theta) -\text{So } x&=\frac{b}{a}\tan(\theta) + \text{So } x&=\frac{b}{a}\tan(\theta) -\text{Let } a^2x^2-b^2&=b^2\sec^2(\theta)-b^2=b^2\tan^2(\theta) + \text{Let } a^2x^2-b^2&=b^2\sec^2(\theta)-b^2=b^2\tan^2(\theta) -\text{So } x&=\frac{b}{a}\sec(\theta) + \text{So } x&=\frac{b}{a}\sec(\theta)

    @@ -201,76 +298,146 @@ trigonometric functions to take advantage of trigonometric identities and simpli Complete the following trigonometric substitution to find \displaystyle\int\dfrac{3}{4+25x^2}\,dx. -\text{Let }4+25x^2&=2+\unknown\theta=\unknown\theta + \text{Let }4+25x^2&=2+\unknown\theta=\unknown\theta -25x^2&=\unknown + 25x^2&=\unknown -x&=\unknown + x&=\unknown -dx&=\unknown\,d\theta + dx&=\unknown\,d\theta -\theta&=\unknown + \theta&=\unknown -\int\frac{3}{4+25x^2}\,dx &=\int\dfrac{3}{\unknown}\,(\unknown\,d\theta) + \int\frac{3}{4+25x^2}\,dx &=\int\dfrac{3}{\unknown}\,(\unknown\,d\theta) -&= \int \unknown \, d\theta + &= \int \unknown \, d\theta -&= \unknown + C + &= \unknown + C -&= \dfrac{3}{10}\arctan\left(\dfrac{5}{2}x\right) + C + &= \dfrac{3}{10}\arctan\left(\dfrac{5}{2}x\right) + C

    + +

    + + + \text{Let }4+25x^2&=4+4\tan^2\theta=4\sec^2\theta + + + 25x^2&=4\tan\theta + + + x&=\dfrac{2}{5}\tan\theta + + + dx&=\dfrac{2}{5}\sec^2\theta\,d\theta + + + \theta&=\arctan\left(\dfrac{5}{2}x\right) + + + + + \int\frac{3}{4+25x^2}\,dx &=\int\dfrac{3}{4\sec^2\theta}\,\left(\dfrac{2}{5}\sec^2\theta\,d\theta\right) + + + &= \int \dfrac{3}{10} \, d\theta + + + &= \dfrac{3}{10}\theta + C + + + &= \dfrac{3}{10}\arctan\left(\dfrac{5}{2}x\right) + C + + +

    +

    -Complete the following trigonometric substitution to find \displaystyle\int\dfrac{7}{x\sqrt{9x^2-16}}\,dx. + Complete the following trigonometric substitution to find \displaystyle\int\dfrac{7}{x\sqrt{9x^2-16}}\,dx. -\text{Let }9x^2-16&=\unknown\theta-16=\unknown\theta + \text{Let }9x^2-16&=\unknown\theta-16=\unknown\theta -9x^2&=\unknown + 9x^2&=\unknown -x&=\unknown + x&=\unknown -dx&=\unknown\,d\theta + dx&=\unknown\,d\theta -\theta&=\unknown + \theta&=\unknown -\displaystyle \int\dfrac{7}{x\sqrt{9x^2-16}}\,dx &=\int\dfrac{7}{\unknown\sqrt{\unknown}}\,(\unknown\,d\theta) + \displaystyle \int\dfrac{7}{x\sqrt{9x^2-16}}\,dx &=\int\dfrac{7}{\unknown\sqrt{\unknown}}\,(\unknown\,d\theta) -&= \int \unknown\, d\theta + &= \int \unknown\, d\theta -&= \unknown + C + &= \unknown + C -&= \frac{7}{4}\arcsec\left(\dfrac{3}{4}x\right) + C + &= \frac{7}{4}\arcsec\left(\dfrac{3}{4}x\right) + C

    + +

    + + + \text{Let }9x^2-16&=16\sec^2\theta-16=16\tan^2\theta + + + 9x^2&=16\sec^2\theta + + + x&=\dfrac{4}{3}\sec\theta + + + dx&=\dfrac{4}{3}\sec\theta\tan\theta\,d\theta + + + \theta&=\arcsec\left(\dfrac{3}{4}x\right) + + + + + \displaystyle \int\dfrac{7}{x\sqrt{9x^2-16}}\,dx &=\int\dfrac{7}{(4/3)\sec\theta\sqrt{16\tan^2\theta}}\,\left(\dfrac{4}{3}\sec\theta\tan\theta\,d\theta\right) + + + &= \int \dfrac{7}{4}\, d\theta + + + &= \unknown + C + + + &= \frac{7}{4}\arcsec\left(\dfrac{3}{4}x\right) + C + + +

    +     @@ -281,56 +448,71 @@ dx&=\unknown\,d\theta

    -\int \frac{\sqrt{-9 \, x^{2} + 16}}{x^{2}}\,dx &= \cdots + \int \frac{\sqrt{-9 \, x^{2} + 16}}{x^{2}}\,dx &= \cdots -&=\int \frac{3\cos^2\theta}{\sin^2\theta}\,d\theta + &=\int \frac{3\cos^2\theta}{\sin^2\theta}\,d\theta -&=-3\theta-3\frac{\cos\theta}{\sin\theta}+C + &=-3\theta-3\frac{\cos\theta}{\sin\theta}+C -&= - 3 \, \arcsin\left(\unknown\right)-\frac{\sqrt{\unknown}}{\unknown} +C + &= - 3 \, \arcsin\left(\unknown\right)-\frac{\sqrt{\unknown}}{\unknown} +C

    + +

    + \displaystyle \int \dfrac{\sqrt{16-9x^2}}{x^2}\, dx = 6\arctan\left(\dfrac{\sqrt{16-9x^2}}{3x+4}\right) - \dfrac{\sqrt{16-9x^2}}{x}+C +

    +

    -\int \frac{2 \, \sqrt{9 \, x^{2} - 16}}{x}\,dx &= \cdots + \int \frac{2 \, \sqrt{9 \, x^{2} - 16}}{x}\,dx &= \cdots -&=\int 8\tan^2\theta\,d\theta + &=\int 8\tan^2\theta\,d\theta -&=8\tan\theta-8\theta+C + &=8\tan\theta-8\theta+C -&= \unknown \, \sqrt{\unknown} - 8 \, \operatorname{arcsec}\left(\unknown\right) +C + &= \unknown \, \sqrt{\unknown} - 8 \, \operatorname{arcsec}\left(\unknown\right) +C

    + +

    + \displaystyle \int \dfrac{2\sqrt{9x^2-16}}{x}\, dx = 2\sqrt{9x^2-16} - 8\arctan\left(\dfrac{\sqrt{9x^2-16}}{4}\right) +

    +

    -\int \frac{1}{\sqrt{81 \, x^{2} + 4}}\,dx&= \cdots + \int \frac{1}{\sqrt{81 \, x^{2} + 4}}\,dx&= \cdots -&=\int \frac{1}{9}\sec\theta\,d\theta + &=\int \frac{1}{9}\sec\theta\,d\theta -&=\frac{1}{9}\log|\sec\theta+\tan\theta|+C + &=\frac{1}{9}\log|\sec\theta+\tan\theta|+C -&= \frac{1}{9} \, \log\left| \unknown + \frac{1}{2} \, \sqrt{\unknown} \right| +C + &= \frac{1}{9} \, \log\left| \unknown + \frac{1}{2} \, \sqrt{\unknown} \right| +C

    + +

    + \displaystyle \int \dfrac{1}{\sqrt{81x^2+4}}\, dx = -\dfrac{1}{9}\ln \left|\sqrt{81x^2+4}-9x\right| + C +

    +     @@ -340,19 +522,38 @@ dx&=\unknown\,d\theta Consider the unit circle x^2+y^2=1. Find a function f(x) so that y=f(x) is the graph of the upper-half semicircle of the unit circle.

    + +

    + y = \sqrt{1-x^2} +

    +     -

    - Find the area under the curve y=f(x) from . -

    + +

    + Find the area under the curve y=f(x) from . +

    +     + +

    + \dfrac{\pi}{2} +

    +     -

    - How does this value compare to what we know about areas of circles? -

    + +

    + How does this value compare to what we know about areas of circles? +

    +     + +

    + The two values are the same. +

    +