AP Calculus AB Applications of Integration — Worked Answer Explanations

Unit 8 · 12% of the AP exam · 12 questions explained

1. Question 1 · Easy

   Find the area between $y=x^2$ and the $x$-axis on $[0,3]$.

   • A
     $3$

     Why not A: Evaluated $f(3)=9$ and divided by $3$ incorrectly.
   • B
     $9$Correct
   • C
     $27$

     Why not C: Computed $[x^3{]}_0^3=27$ without dividing by $3$.
   • D
     $6$

     Why not D: Integrated $2x$ instead of $x^2$.
   Explanation

   ${\int }_0^3{x}^{2}dx={\left[\frac{x^3}{3}\right]}_0^3=\frac{27}{3}-0=9$.

   Key takeaway

   Area under a curve on $[a,b]$ is $\int_a^b f(x)\,dx$ — apply the power rule for integration and evaluate the antiderivative.

2. Question 2 · Easy

   What is the average value of $f(x)=4x$ on $[1,3]$?

   • A
     $4$

     Why not A: Evaluated $f$ at the midpoint $x=2$ but forgot the factor from the formula.
   • B
     $8$Correct
   • C
     $16$

     Why not C: Computed ${\int }_1^{3}4xdx=16$ without dividing by $(b-a)=2$.
   • D
     $12$

     Why not D: Evaluated $f(3)=12$ as the "average" without integrating.
   Explanation

   Average value: $\frac{1}{b-a}{\int }_a^{b}f(x)dx=\frac{1}{2}{\int }_1^{3}4xdx=\frac{1}{2}{\left[2x^2\right]}_1^3=\frac{1}{2}(18-2)=8$.

   Key takeaway

   Average value formula: $f_{\text{avg}} = \dfrac{1}{b-a}\int_a^b f(x)\,dx$ — divide the integral by the interval length.

3. Question 3 · Easy

   A particle moves along the $x$-axis with velocity $v(t)=t^2-4$ for $0\le t\le 3$. What is the total displacement?

   • A
     $-3$Correct
   • B
     $3$

     Why not B: Computed absolute value of net displacement, getting the sign wrong.
   • C
     $\frac{11}{3}$

     Why not C: Computed total distance (using $|v|$) rather than displacement.
   • D
     $0$

     Why not D: Assumed the particle returns to start because $v$ changes sign.
   Explanation

   Displacement $={\int }_0^3(t^2-4)dt={\left[\frac{t^3}{3}-4t\right]}_0^3=(9-12)-(0)=-3$.

   Key takeaway

   Displacement is $\int_a^b v(t)\,dt$; negative values indicate net leftward/downward motion.

4. Question 4 · Easy

   Find the area enclosed between $y=x$ and $y=x^2$ on $[0,1]$.

   • A
     $\frac{1}{3}$

     Why not A: Integrated ${\int }_0^1{x}^{2}dx$ instead of the difference $x-x^2$.
   • B
     $\frac{1}{6}$Correct
   • C
     $\frac{1}{2}$

     Why not C: Integrated only ${\int }_0^{1}xdx$ without subtracting $x^2$.
   • D
     $1$

     Why not D: Added the two integrals instead of subtracting.
   Explanation

   On $[0,1]$, $x\ge x^2$. Area $={\int }_0^1(x-x^2)dx={\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$.

   Key takeaway

   Area between curves: integrate (top function $-$ bottom function) over the interval; identify which curve is on top first.

5. Question 5 · Medium

   Find the area between $y=\sin x$ and $y=\cos x$ on $\left[\frac{\pi }{4},\frac{5\pi }{4}\right]$.

   • A
     $0$

     Why not A: Integrated $\sin x-\cos x$ without noting $\sin x\ge \cos x$ throughout the interval.
   • B
     $\sqrt{2}$

     Why not B: Only evaluated the antiderivative at one endpoint.
   • C
     $2\sqrt{2}$Correct
   • D
     $4$

     Why not D: Used an incorrect antiderivative.
   Explanation

   On $[\pi /4,5\pi /4]$, $\sin x\ge \cos x$. Area $={\int }_{\pi /4}^{5\pi /4}(\sin x-\cos x)dx=[-\cos x-\sin x{]}_{\pi /4}^{5\pi /4}$. At $5\pi /4$: $-\cos (5\pi /4)-\sin (5\pi /4)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$. At $\pi /4$: $-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}=-\sqrt{2}$. Area $=\sqrt{2}-(-\sqrt{2})=2\sqrt{2}$.

   Key takeaway

   Area between trig curves: determine which function is on top over the interval, integrate the difference, evaluate carefully at the bounds.

6. Question 6 · Medium

   Using the disk method, find the volume of the solid formed by rotating $y=\sqrt{x}$ on $[0,4]$ about the $x$-axis.

   • A
     $4\pi$

     Why not A: Forgot to square $R(x)=\sqrt{x}$ in the disk formula.
   • B
     $8\pi$Correct
   • C
     $16\pi$

     Why not C: Used $R(x)=x$ instead of $R(x)=\sqrt{x}$.
   • D
     $2\pi$

     Why not D: Divided by $2$ after integrating, confusing with the average value formula.
   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 (rotation about $x$-axis): $V = \pi\int_a^b [R(x)]^2\,dx$ where $R(x)$ is the radius (the function value).

7. Question 7 · Medium

   The acceleration of a particle is $a(t)=6t$ with $v(0)=-2$ and $s(0)=4$. Find $s(3)$.

   • A
     $27$

     Why not A: Computed ${\int }_0^{3}a(t)dt$ without initial conditions.
   • B
     $25$Correct
   • C
     $31$

     Why not C: Added $s(0)=4$ to ${\int }_0^{3}v(t)dt$ without incorporating $v(0)=-2$.
   • D
     $22$

     Why not D: Sign error: used $v(0)=+2$ rather than $-2$.
   Explanation

   Integrate $a$: $v(t)=3t^2+C_1$. Apply $v(0)=-2$: $C_1=-2$, so $v(t)=3t^2-2$. Integrate $v$: $s(t)=t^3-2t+C_2$. Apply $s(0)=4$: $C_2=4$. $s(3)=27-6+4=25$.

   Key takeaway

   Work from acceleration to position by integrating twice; apply each initial condition after each integration step.

8. Question 8 · Medium

   Find the volume of the solid with square cross-sections perpendicular to the $x$-axis, where the base is the region bounded by $y=\sqrt{x}$ and $y=0$ on $[0,4]$.

   • A
     $4$

     Why not A: Integrated $\sqrt{x}$ rather than $x$ (the square of the side length).
   • B
     $8$Correct
   • C
     $8\pi$

     Why not C: Used the disk formula $\pi \int R^2$ instead of the cross-section formula $\int A(x)$.
   • D
     $16$

     Why not D: Computed ${\int }_0^{4}xdx$ without upper limit adjustment.
   Explanation

   Each cross-section is a square with side length $s(x)=\sqrt{x}$, so area $A(x)=(\sqrt{x}{)}^2=x$. Volume $={\int }_0^{4}xdx={\left[\frac{x^2}{2}\right]}_0^4=8$.

   Key takeaway

   Cross-section method: $V = \int A(x)\,dx$ where $A(x)$ is the area of the cross-section at position $x$.

9. Question 9 · Medium

   A tank contains $100$ gallons of water. Water drains at rate $r(t)=5t$ gal/min. How much water drains out in the first $4$ minutes?

   • A
     $20$ gallons

     Why not A: Computed $r(4)=20$ (the instantaneous rate) rather than integrating.
   • B
     $40$ gallonsCorrect
   • C
     $80$ gallons

     Why not C: Used $5\times 4^2$ without dividing by $2$.
   • D
     $10$ gallons

     Why not D: Divided by $2$ incorrectly: got $5t/2$ at $t=4$ without proper integration limits.
   Explanation

   Total amount drained $={\int }_0^{4}5tdt={\left[\frac{5t^2}{2}\right]}_0^4=\frac{5\cdot 16}{2}=40$ gallons.

   Key takeaway

   Accumulation problems: integrate the rate function over the time interval to find the total amount accumulated or drained.

10. Question 10 · Hard

    Use the washer method to find the volume when the region between $y=x$ and $y=x^2$ (on $[0,1]$) is rotated about the $x$-axis.

    • A
      $\frac{\pi }{6}$

      Why not A: Computed the area of the 2D region, not the volume of revolution.
    • B
      $\frac{2\pi }{15}$Correct
    • C
      $\frac{\pi }{5}$

      Why not C: Used the outer radius only, omitting the inner radius in the washer formula.
    • D
      $\frac{\pi }{15}$

      Why not D: Subtracted the integrals before squaring: $(x-x^2{)}^2$ instead of $(x^2-x^4)$.
    Explanation

    Washer method: $V=\pi {\int }_0^1[R^2(x)-r^2(x)]dx$ where $R(x)=x$ (outer) and $r(x)=x^2$ (inner). $V=\pi {\int }_0^1(x^2-x^4)dx=\pi {\left[\frac{x^3}{3}-\frac{x^5}{5}\right]}_0^1=\pi \left(\frac{1}{3}-\frac{1}{5}\right)=\frac{2\pi }{15}$.

    Key takeaway

    Washer method: subtract the squared inner radius from the squared outer radius inside the integral — do not subtract before squaring.

11. Question 11 · Hard

    A particle's velocity is $v(t)=t^2-4t+3$ for $0\le t\le 4$. What is the total distance traveled?

    • A
      $\frac{4}{3}$

      Why not A: Computed net displacement ${\int }_0^{4}vdt$ without splitting at zeros of $v$.
    • B
      $4$Correct
    • C
      $\frac{16}{3}$

      Why not C: Added the magnitudes of the two sub-interval integrals incorrectly.
    • D
      $\frac{8}{3}$

      Why not D: Only computed the integral on one sub-interval.
    Explanation

    $v(t)=(t-1)(t-3)$; $v=0$ at $t=1$ and $t=3$. On $[0,1]$: $v\ge 0$; on $[1,3]$: $v\le 0$; on $[3,4]$: $v\ge 0$. Distance =∫01​vdt+​∫13​vdt​+∫34​vdt. ${\int }_0^1(t^2-4t+3)dt=[t^3/3-2t^2+3t{]}_0^1=1/3-2+3=4/3$. ${\int }_1^3(t^2-4t+3)dt=[t^3/3-2t^2+3t{]}_1^3=(9-18+9)-(1/3-2+3)=0-4/3=-4/3$, so $|-4/3|=4/3$. ${\int }_3^4(t^2-4t+3)dt=[t^3/3-2t^2+3t{]}_3^4=(64/3-32+12)-(9-18+9)=4/3-0=4/3$. Total $=4/3+4/3+4/3=4$.

    Key takeaway

    Total distance uses $\int |v(t)|\,dt$; split the integral at zeros of $v$ and add absolute values of each piece.

12. Question 12 · Hard

    Find the volume of the solid formed when the region bounded by $y=e^x$, the $x$-axis, $x=0$, and $x=1$ is rotated about the $x$-axis.

    • A
      $\pi (e-1)$

      Why not A: Integrated $e^x$ instead of $e^{2x}$ (forgot to square the radius).
    • B
      $\frac{\pi (e^2-1)}{2}$Correct
    • C
      $\pi e^2$

      Why not C: Evaluated $\pi [e^x{]}_0^1$ without the factor of $1/2$ from integrating $e^{2x}$.
    • D
      $e^2-1$

      Why not D: Forgot to multiply by $\pi$ in the disk formula.
    Explanation

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

    Key takeaway

    Disk method with exponentials: square the function first to get $e^{2x}$, then integrate $\int e^{2x}\,dx = e^{2x}/2$.