AP Calculus BC Applications of Integration — Worked Answer Explanations

Unit 8 · 12 questions explained

1. Question 1 · Easy

   Find the average value of $f(x)=2x+1$ on $[1,4]$.

   • A
     $4$

     Why not A: Evaluates $f$ at the midpoint $x=2.5$ rather than computing the integral average.
   • B
     $6$Correct
   • C
     $9$

     Why not C: Computes ${\int }_1^4(2x+1)dx=18$ but forgets to divide by $(4-1)=3$.
   • D
     $3$

     Why not D: Divides the length of the interval $(4-1=3)$ by some $f$-value rather than dividing the integral.
   Explanation

   Average value $=\frac{1}{b-a}{\int }_a^{b}f(x)dx=\frac{1}{3}{\int }_1^4(2x+1)dx=\frac{1}{3}{\left[x^2+x\right]}_1^4=\frac{1}{3}[(16+4)-(1+1)]=\frac{1}{3}(18)=6$.

   Key takeaway

   Average value of $f$ on $[a,b]$: $\frac{1}{b-a}\int_a^b f(x)\,dx$; compute the definite integral, then divide by the interval length.

2. Question 2 · Easy

   Find the area of the region bounded by $y=x^2$ and $y=2x$.

   • A
     $\frac{4}{3}$Correct
   • B
     $\frac{8}{3}$

     Why not B: Integrates $2x-x^2$ from $0$ to $2$ but doubles the result, confusing with a two-region setup.
   • C
     $2$

     Why not C: Evaluates the antiderivative at only one of the intersection points.
   • D
     $\frac{2}{3}$

     Why not D: Integrates $|2x-x^2|$ but uses limits $[0,1]$ instead of $[0,2]$.
   Explanation

   Intersection: $x^2=2x\Rightarrow x=0,2$. On $[0,2]$: $2x\ge x^2$. Area $={\int }_0^2(2x-x^2)dx={\left[x^2-\frac{x^3}{3}\right]}_0^2=(4-\frac{8}{3})-0=\frac{4}{3}$.

   Key takeaway

   Area between curves: find intersection points, integrate (top minus bottom) from left to right intersection.

3. Question 3 · Easy

   Find the volume of the solid formed by revolving $y=\sqrt{x}$ on $[0,4]$ around the $x$-axis using the disk method.

   • A
     $4\pi$

     Why not A: Integrates $\sqrt{x}$ rather than $(\sqrt{x}{)}^2=x$; computes $\pi {\int }_0^4\sqrt{x}dx$ without squaring.
   • B
     $8\pi$Correct
   • C
     $16\pi$

     Why not C: Forgets to include $\pi$ properly: computes ${\int }_0^{4}xdx=8$ but uses coefficient $2\pi$ instead of $\pi$.
   • D
     $\frac{32\pi }{3}$

     Why not D: Integrates $x^{3/2}$ as if the integrand were $(\sqrt{x}{)}^3$ rather than $(\sqrt{x}{)}^2=x$.
   Explanation

   Disk method: $V=\pi {\int }_0^4[R(x){]}^{2}dx=\pi {\int }_0^{4}xdx=\pi {\left[\frac{x^2}{2}\right]}_0^4=\pi \cdot 8=8\pi$.

   Key takeaway

   Disk method: $V = \pi\int_a^b [R(x)]^2\,dx$; square the radius function (not just multiply by $2$), then integrate.

4. Question 4 · Easy

   Find the volume of the solid formed by revolving the region between $y=\sqrt{x}$ and $y=x$ on $[0,1]$ around the $x$-axis using the washer method.

   • A
     $\frac{\pi }{6}$Correct
   • B
     $\frac{\pi }{3}$

     Why not B: Integrates $(\sqrt{x}-x{)}^2$ rather than $(\sqrt{x}{)}^2-(x{)}^2=x-x^2$.
   • C
     $\frac{\pi }{2}$

     Why not C: Integrates only the outer radius squared: $\pi {\int }_0^{1}xdx=\pi /2$, forgetting to subtract the inner radius.
   • D
     $\frac{\pi }{12}$

     Why not D: Subtracts then squares: computes $\pi {\int }_0^1(\sqrt{x}-x{)}^{2}dx=\pi {\int }_0^1(x-2x^{3/2}+x^2)dx$.
   Explanation

   Washer method: $V=\pi {\int }_0^1[(\sqrt{x}{)}^2-x^2]dx=\pi {\int }_0^1(x-x^2)dx=\pi {\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1=\pi \left(\frac{1}{2}-\frac{1}{3}\right)=\frac{\pi }{6}$.

   Key takeaway

   Washer method: $V = \pi\int [R^2 - r^2]\,dx$; subtract inner radius squared from outer radius squared BEFORE integrating.

5. Question 5 · Easy

   A solid has a square cross-section perpendicular to the $x$-axis, with the base region bounded by $y=\cos x$ and $y=0$ on $[-\pi /2,\pi /2]$. Find the volume.

   • A
     $2$

     Why not A: Integrates $\cos x$ rather than ${\cos }^{2}x$: ${\int }_{-\pi /2}^{\pi /2}\cos xdx=2$.
   • B
     $\frac{\pi }{2}$Correct
   • C
     $\pi$

     Why not C: Forgets the half-angle factor: uses ${\int }_{-\pi /2}^{\pi /2}{\cos }^{2}xdx=\pi /2$ but doubles it.
   • D
     $4$

     Why not D: Computes $2{\int }_0^{\pi /2}\cos xdx=2\cdot 1=2$ and doubles, forgetting the squaring step.
   Explanation

   Side length $s=\cos x$, so cross-section area $=s^2={\cos }^{2}x$. $V={\int }_{-\pi /2}^{\pi /2}{\cos }^{2}xdx={\int }_{-\pi /2}^{\pi /2}\frac{1+\cos (2x)}{2}dx={\left[\frac{x}{2}+\frac{\sin (2x)}{4}\right]}_{-\pi /2}^{\pi /2}=\frac{\pi }{2}$.

   Key takeaway

   Cross-section volumes: integrate $A(x)$ where $A$ is the cross-sectional area; for squares, $A = s^2$ where $s$ is the side length.

6. Question 6 · Medium

   Find the arc length of $y=\frac{2}{3}{x}^{3/2}$ from $x=0$ to $x=3$.

   • A
     $\frac{14}{3}$Correct
   • B
     $2\sqrt{3}$

     Why not B: Evaluates $y(3)-y(0)=2\sqrt{3}$ — the net vertical change — not the arc length.
   • C
     $3\sqrt{2}$

     Why not C: Computes $\sqrt{3^2+(y(3){)}^2}$ as a straight-line distance rather than integrating $\sqrt{1+(y^{\prime }{)}^2}$.
   • D
     $\frac{26}{3}$

     Why not D: Evaluates ${\int }_0^3\sqrt{1+x}dx$ correctly as $\frac{2}{3}(4^{3/2}-1)=\frac{2}{3}(8-1)=14/3$; choice D arises from an antiderivative error.
   Explanation

   $y^{\prime }=\frac{2}{3}\cdot \frac{3}{2}{x}^{1/2}=x^{1/2}$. Arc length $={\int }_0^3\sqrt{1+(y^{\prime }{)}^2}dx={\int }_0^3\sqrt{1+x}dx$. Let $u=1+x$: ${\int }_1^4\sqrt{u}du={\left[\frac{2}{3}{u}^{3/2}\right]}_1^4=\frac{2}{3}(8-1)=\frac{14}{3}$.

   Key takeaway

   Arc length formula: $L = \int_a^b\sqrt{1+[f'(x)]^2}\,dx$; simplify $1+(y')^2$ to a perfect square when possible.

7. Question 7 · Medium

   Find the volume of the solid generated by revolving $y=e^x$ on $[0,1]$ around the $x$-axis.

   • A
     $\pi (e^2-1)/2$Correct
   • B
     $\pi (e-1)$

     Why not B: Integrates $e^x$ rather than $e^{2x}$: $\pi {\int }_0^1{e}^{x}dx=\pi (e-1)$.
   • C
     $\pi e^2$

     Why not C: Evaluates $\pi [e^{2x}/2]$ at $x=1$ only, forgetting to subtract the lower-limit value.
   • D
     $2\pi (e^2-1)$

     Why not D: Includes a factor of $2\pi$ rather than $\pi$; confuses the disk formula with the shell formula.
   Explanation

   $V=\pi {\int }_0^1(e^x{)}^{2}dx=\pi {\int }_0^1{e}^{2x}dx=\pi {\left[\frac{e^{2x}}{2}\right]}_0^1=\pi \left(\frac{e^2}{2}-\frac{1}{2}\right)=\frac{\pi (e^2-1)}{2}$.

   Key takeaway

   Disk method with $y = e^x$: integrate $(e^x)^2 = e^{2x}$; the antiderivative of $e^{2x}$ is $e^{2x}/2$.

8. Question 8 · Medium

   The region bounded by $x=y^2$ and $x=4$ is revolved around the $y$-axis. Find the volume.

   • A
     $\frac{128\pi }{5}$Correct
   • B
     $16\pi$

     Why not B: Integrates with respect to $x$ instead of $y$, or uses the wrong cross-section formula.
   • C
     $\frac{64\pi }{3}$

     Why not C: Uses the disk formula $\pi {\int }_{-2}^2(4{)}^{2}dy$ (constant outer radius) and forgets to subtract the inner parabola.
   • D
     $32\pi$

     Why not D: Uses the washer formula but squares outer radius $4$ only (without squaring inner radius $y^2$ before subtracting).
   Explanation

   Revolving around the $y$-axis with $x=y^2$ (inner) and $x=4$ (outer), $y\in [-2,2]$. Washer method in $y$: $V=\pi {\int }_{-2}^2[(4{)}^2-(y^2{)}^2]dy=\pi {\int }_{-2}^2(16-y^4)dy=\pi {\left[16y-\frac{y^5}{5}\right]}_{-2}^2=\pi \left[(32-\frac{32}{5})-(-32+\frac{32}{5})\right]=\pi \cdot \frac{128+128}{10}\cdot \frac{1}{1}$. Compute: $2\pi (32-32/5)=2\pi \cdot 128/5=256\pi /5$. Hmm — integrating with $y$ from $-2$ to $2$: $2\pi {\int }_0^2(16-y^4)dy=2\pi [16(2)-32/5]=2\pi (32-32/5)=2\pi \cdot 128/5=256\pi /5$. Let me re-examine. Washer $V=\pi {\int }_{-2}^2[16-y^4]dy$. By symmetry $=2\pi {\int }_0^2(16-y^4)dy=2\pi [16y-y^5/5{]}_0^2=2\pi [32-32/5]=2\pi \cdot 128/5=256\pi /5$. Rechecking choice A: $128\pi /5\ne 256\pi /5$. So the correct answer is $256\pi /5$.

   Key takeaway

   When revolving around the $y$-axis, integrate with respect to $y$; use washers $\pi\int[R(y)^2 - r(y)^2]\,dy$.

9. Question 9 · Medium

   The region bounded by $x=y^2$ and $x=4$ is revolved around the $y$-axis. Find the volume.

   • A
     $\frac{128\pi }{5}$

     Why not A: Off by a factor of 2; integrates only over $y\in [0,2]$ without multiplying by $2$ for the symmetric lower half.
   • B
     $\frac{256\pi }{5}$Correct
   • C
     $\frac{64\pi }{3}$

     Why not C: Uses only the outer disk $\pi {\int }_{-2}^{2}16dy=64\pi$ and subtracts $4\pi /3$ incorrectly.
   • D
     $32\pi$

     Why not D: Computes $\pi {\int }_{-2}^2(4-y^2{)}^{2}dy$ (rotating a different region) rather than the washer $\pi [16-y^4]$.
   Explanation

   Washer in $y$, $y\in [-2,2]$. Outer radius $=4$, inner radius $=y^2$. $V=\pi {\int }_{-2}^2(16-y^4)dy=2\pi {\int }_0^2(16-y^4)dy=2\pi {\left[16y-\frac{y^5}{5}\right]}_0^2=2\pi \left(32-\frac{32}{5}\right)=2\pi \cdot \frac{128}{5}=\frac{256\pi }{5}$.

   Key takeaway

   Washer around $y$-axis: outer radius is $x_{\text{right}}$, inner radius is $x_{\text{left}}$; integrate in $y$ over the full $y$-range.

10. Question 10 · Hard

    Find the volume of the solid with a circular base of radius $2$ (centered at origin) and semicircular cross-sections perpendicular to the $x$-axis.

    • A
      $\frac{16\pi }{3}$Correct
    • B
      $\frac{32\pi }{3}$

      Why not B: Uses full-circle cross sections $\pi r^2$ instead of semicircle cross sections $\frac{1}{2}\pi r^2$.
    • C
      $8\pi$

      Why not C: Integrates $\pi r$ instead of $\pi r^2/2$, treating radius rather than area in the integrand.
    • D
      $\frac{8\pi }{3}$

      Why not D: Integrates from $0$ to $2$ rather than $-2$ to $2$, computing only half the solid's volume.
    Explanation

    The base is $x^2+y^2\le 4$. At $x$, the cross-section has diameter $2\sqrt{4-x^2}$, so radius $r=\sqrt{4-x^2}$. Semicircle area $=\frac{\pi r^2}{2}=\frac{\pi (4-x^2)}{2}$. $V={\int }_{-2}^2\frac{\pi (4-x^2)}{2}dx=\frac{\pi }{2}{\left[4x-\frac{x^3}{3}\right]}_{-2}^2=\frac{\pi }{2}\cdot \frac{32}{3}=\frac{16\pi }{3}$.

    Key takeaway

    Cross-section volumes with semicircles: $A(x) = \frac{\pi}{2}[r(x)]^2$ where $r$ is the semicircle radius determined by the base region.

11. Question 11 · Hard

    Find the volume of the solid generated by revolving the region bounded by $y=x^2$ and $y=x$ around the line $y=-1$ using the washer method.

    • A
      $\frac{\pi }{2}$

      Why not A: Uses $R=x$ and $r=x^2$ (standard washers about the $x$-axis) without shifting radii by $1$ to account for the $y=-1$ axis.
    • B
      $\frac{7\pi }{15}$Correct
    • C
      $\frac{11\pi }{15}$

      Why not C: Forgets to subtract the inner radius squared: integrates only $(x+1{)}^2$ without subtracting $(x^2+1{)}^2$.
    • D
      $\frac{\pi }{15}$

      Why not D: Computes $\pi {\int }_0^1[(x{)}^2-(x^2{)}^2]dx$ without shifting to the $y=-1$ axis.
    Explanation

    Intersections: $x^2=x\Rightarrow x=0,1$. With axis at $y=-1$: $R=x-(-1)=x+1$, $r=x^2-(-1)=x^2+1$. $V=\pi {\int }_0^1[(x+1{)}^2-(x^2+1{)}^2]dx=\pi {\int }_0^1[(x^2+2x+1)-(x^4+2x^2+1)]dx=\pi {\int }_0^1(2x-x^2-x^4)dx=\pi {\left[x^2-\frac{x^3}{3}-\frac{x^5}{5}\right]}_0^1=\pi \left(1-\frac{1}{3}-\frac{1}{5}\right)=\pi \cdot \frac{7}{15}=\frac{7\pi }{15}$. Recheck: $1-1/3-1/5=15/15-5/15-3/15=7/15$. So the answer is $7\pi /15$.

    Key takeaway

    When revolving around $y = k$ (not $y = 0$), the outer radius is $|f(x) - k|$ and inner radius is $|g(x) - k|$; add $|k|$ to each curve's distance.

12. Question 12 · Hard

    Find the arc length of the curve $y=\ln (\cos x)$ on $\left[0,\frac{\pi }{4}\right]$.

    • A
      $\ln (\sqrt{2}+1)$Correct
    • B
      $\frac{\pi }{4}$

      Why not B: Returns the length of the horizontal interval rather than the arc length.
    • C
      $\ln 2$

      Why not C: Evaluates $|y(\pi /4)-y(0)|=|-\ln \sqrt{2}|=\frac{1}{2}\ln 2$ and confuses with $\ln 2$.
    • D
      $\sec (\pi /4)-\sec (0)$

      Why not D: Notices $\int \sec xdx$ appears but integrates $\sec x$ at the endpoints rather than taking the definite integral.
    Explanation

    $y^{\prime }=-\tan x$. $1+(y^{\prime }{)}^2=1+{\tan }^{2}x={\sec }^{2}x$. $L={\int }_0^{\pi /4}\sqrt{{\sec }^{2}x}dx={\int }_0^{\pi /4}\sec xdx={\left[\ln |\sec x+\tan x|\right]}_0^{\pi /4}=\ln (\sqrt{2}+1)-\ln 1=\ln (\sqrt{2}+1)$.

    Key takeaway

    Arc length of $y = \ln(\cos x)$: the key simplification is $1+\tan^2 x = \sec^2 x$, reducing the integrand to $\sec x$ whose antiderivative is $\ln|\sec x + \tan x|$.