AP Calculus BC Integration and Accumulation of Change — Worked Answer Explanations

Unit 6 · 12 questions explained

1. Question 1 · Easy

   Evaluate $\int (3x^2-4x+5)dx$.

   • A
     $6x-4+C$

     Why not A: Differentiates rather than integrates the polynomial.
   • B
     $x^3-2x^2+5x$

     Why not B: Correct antiderivative terms but omits the constant of integration $C$.
   • C
     $x^3-2x^2+5x+C$Correct
   • D
     $x^3-4x^2+5x+C$

     Why not D: Integrates $-4x$ as $-4x^2$ instead of $-2x^2$ (forgets to divide by the new exponent $2$).
   Explanation

   Apply the power rule for integrals term by term: $\int 3x^{2}dx=x^3$, $\int 4xdx=2x^2$, $\int 5dx=5x$. Result: $x^3-2x^2+5x+C$.

   Key takeaway

   Antidifferentiate polynomial terms with the power rule $\int x^n\,dx = x^{n+1}/(n+1)$; always include the constant $C$.

2. Question 2 · Easy

   Evaluate ${\int }_1^4\sqrt{x}dx$.

   • A
     $\frac{14}{3}$Correct
   • B
     $\frac{4}{3}$

     Why not B: Evaluates ${\int }_1^4{x}^{1/2}dx$ by using exponent $\frac{1}{2}$ but divides by $\frac{1}{2}$ instead of $\frac{3}{2}$.
   • C
     $\frac{2}{3}(8-1)=\frac{14}{3}$... wait: $3$

     Why not C: Computes $\frac{2}{3}[x^{3/2}{]}_1^4=\frac{2}{3}(8-1)$ but truncates to $3$.
   • D
     $\frac{7}{3}$

     Why not D: Computes $\frac{2}{3}(4-1)=2$ using $x^1$ instead of $x^{3/2}$ in the antiderivative.
   Explanation

   ${\int }_1^4{x}^{1/2}dx={\left[\frac{x^{3/2}}{3/2}\right]}_1^4=\frac{2}{3}{\left[x^{3/2}\right]}_1^4=\frac{2}{3}(4^{3/2}-1^{3/2})=\frac{2}{3}(8-1)=\frac{14}{3}$.

   Key takeaway

   For $\int x^{1/2}\,dx$: add $1$ to get $x^{3/2}$ and divide by $3/2$ (equivalently multiply by $2/3$).

3. Question 3 · Easy

   Evaluate $\int \sin x\cos xdx$.

   • A
     $-{\cos }^{2}x+C$

     Why not A: Integrates $\cos x$ alone as $-\sin x$ (wrong sign); this is not the correct antiderivative.
   • B
     $\frac{{\sin }^{2}x}{2}+C$Correct
   • C
     ${\cos }^{2}x+C$

     Why not C: Uses the chain rule (derivative) instead of integration: $\frac{d}{dx}[{\sin }^{2}x/2]=\sin x\cos x$, but selecting ${\cos }^{2}x$ conflates this with the derivative of ${\cos }^{2}x$.
   • D
     ${\sin }^{2}x\cos x+C$

     Why not D: Applies a product rule pattern rather than substitution.
   Explanation

   $u$-substitution: $u=\sin x$, $du=\cos xdx$. $\int \sin x\cos xdx=\int udu=\frac{u^2}{2}+C=\frac{{\sin }^{2}x}{2}+C$. (Equivalently, $\int \sin x\cos xdx=\frac{1}{2}\int \sin (2x)dx=-\frac{1}{4}\cos (2x)+C$, which differs by a constant.)

   Key takeaway

   Use $u$-substitution with $u = \sin x$ to integrate $\sin x\cos x$; both approaches ($u$-sub and double angle) give equivalent antiderivatives.

4. Question 4 · Easy

   If $F(x)={\int }_0^x\cos (t^2)dt$, find $F^{\prime }(x)$ by the Fundamental Theorem of Calculus Part 1.

   • A
     $\cos (x)$

     Why not A: Ignores the $t^2$ inside cosine; applies FTC as if the integrand were $\cos (t)$.
   • B
     $\cos (x^2)$Correct
   • C
     $2x\cos (x^2)$

     Why not C: Applies chain rule as if the upper limit $x$ required a factor $1$ (it does), but instead includes a spurious $2x$ factor from the integrand's argument — FTC Part 1 does not differentiate the integrand argument.
   • D
     $\sin (x^2)$

     Why not D: Applies $d/dx[\int \sin tdt]=-\cos t$ pattern incorrectly; FTC Part 1 simply substitutes the upper limit.
   Explanation

   By FTC Part 1: $\frac{d}{dx}{\int }_0^{x}f(t)dt=f(x)$. Here $f(t)=\cos (t^2)$, so $F^{\prime }(x)=\cos (x^2)$. The chain rule would only apply if the upper limit were a function of $x$ other than $x$ itself.

   Key takeaway

   FTC Part 1: $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$; simply replace $t$ with $x$ in the integrand — no further chain rule unless the limit is a composite function.

5. Question 5 · Easy

   Evaluate $\int \frac{2x}{x^2+3}dx$.

   • A
     $\ln |x^2+3|+C$Correct
   • B
     $2\ln |x^2+3|+C$

     Why not B: Integrates $2x/(x^2+3)$ as $2\cdot (1/1)\cdot \ln |x^2+3|$ without accounting for the substitution $du=2xdx$ already in the numerator.
   • C
     $\frac{-2x}{(x^2+3{)}^2}+C$

     Why not C: Differentiates rather than integrates; this is $d/dx[-1/(x^2+3)]$ (off by factor), not the integral.
   • D
     $\frac{1}{2}\ln |x^2+3|+C$

     Why not D: Introduces a spurious factor of $\frac{1}{2}$; no extra factor is needed since $du=2xdx$ matches the numerator exactly.
   Explanation

   $u=x^2+3$, $du=2xdx$. $\int \frac{2x}{x^2+3}dx=\int \frac{du}{u}=\ln |u|+C=\ln |x^2+3|+C$.

   Key takeaway

   Recognize $\int f'/f\,dx = \ln|f| + C$; when the numerator is the exact derivative of the denominator, $u$-substitution immediately gives $\ln$.

6. Question 6 · Easy

   Using the Left Riemann Sum with $n=4$ equal subintervals, approximate ${\int }_0^2{x}^{2}dx$.

   • A
     $\frac{7}{4}$Correct
   • B
     $\frac{15}{4}$

     Why not B: Uses the Right Riemann Sum instead of the Left Riemann Sum.
   • C
     $\frac{8}{3}$

     Why not C: Computes the exact integral ${\int }_0^2{x}^{2}dx=8/3$ rather than the left Riemann sum approximation.
   • D
     $\frac{11}{4}$

     Why not D: Computes the Midpoint Rule sum with $n=4$ rather than the Left Rule.
   Explanation

   $\Delta x=(2-0)/4=0.5$. Left endpoints: $x=0,0.5,1,1.5$. Sum: $\Delta x[f(0)+f(0.5)+f(1)+f(1.5)]=0.5[0+0.25+1+2.25]=0.5\cdot 3.5=\frac{7}{4}$.

   Key takeaway

   Left Riemann Sum: use the left endpoint of each subinterval; multiply each $f$-value by $\Delta x$ and sum.

7. Question 7 · Easy

   Evaluate $\int x{e}^{x^2}dx$.

   • A
     $\frac{1}{2}{e}^{x^2}+C$Correct
   • B
     $e^{x^2}+C$

     Why not B: Integrates $x{e}^{x^2}$ without accounting for the $1/2$ factor from the substitution $du=2xdx$.
   • C
     $x^2{e}^{x^2}/2+C$

     Why not C: Applies integration by parts incorrectly, treating $e^{x^2}$ as if differentiating $x^2$ produces a clean result.
   • D
     $2x{e}^{x^2}+C$

     Why not D: Takes the derivative of $e^{x^2}$ (which is $2x{e}^{x^2}$) rather than the antiderivative.
   Explanation

   $u=x^2$, $du=2xdx$, so $xdx=\frac{1}{2}du$. $\int x{e}^{x^2}dx=\int e^u\cdot \frac{1}{2}du=\frac{1}{2}{e}^u+C=\frac{1}{2}{e}^{x^2}+C$.

   Key takeaway

   For $\int x e^{x^2}\,dx$, use $u = x^2$; the $x\,dx$ in the integrand pairs with $du = 2x\,dx$ giving a factor of $1/2$.

8. Question 8 · Medium

   Evaluate $\int x\cos xdx$ using integration by parts.

   • A
     $x\sin x+\cos x+C$Correct
   • B
     $x\sin x-\cos x+C$

     Why not B: Sign error in the second term: $\int \sin xdx=-\cos x$, so after integration by parts the remaining integral contributes $+\cos x$, not $-\cos x$.
   • C
     $\cos x-x\sin x+C$

     Why not C: Swaps $u$ and $dv$: lets $u=\cos x$ and $dv=xdx$, leading to a more complex integral.
   • D
     $x\cos x+\sin x+C$

     Why not D: Returns a form resembling the derivative of $x\sin x$; confuses differentiation with integration.
   Explanation

   Let $u=x$, $dv=\cos xdx$. Then $du=dx$, $v=\sin x$. $\int x\cos xdx=x\sin x-\int \sin xdx=x\sin x-(-\cos x)+C=x\sin x+\cos x+C$.

   Key takeaway

   Integration by parts: $\int u\,dv = uv - \int v\,du$; choose $u$ as the polynomial and $dv$ as the trig or exponential factor.

9. Question 9 · Medium

   Evaluate ${\int }_0^{\pi }{\sin }^{2}xdx$.

   • A
     $0$

     Why not A: Confuses ${\sin }^{2}x$ with $\sin x$: ${\int }_0^{\pi }\sin xdx=2\ne 0$, but ${\sin }^{2}x\ge 0$ so the integral is positive.
   • B
     $\frac{\pi }{2}$Correct
   • C
     $\pi$

     Why not C: Forgets the factor of $1/2$ from the half-angle identity: ${\int }_0^{\pi }{\sin }^{2}xdx\ne {\int }_0^{\pi }1dx=\pi$.
   • D
     $1$

     Why not D: Evaluates $[-\frac{1}{4}\sin (2x){]}_0^{\pi }$ as $0$ and adds only the $x/2$ piece as $\pi /2$, but then rounds or truncates to $1$.
   Explanation

   Half-angle identity: ${\sin }^{2}x=\frac{1-\cos (2x)}{2}$. ${\int }_0^{\pi }\frac{1-\cos (2x)}{2}dx=\frac{1}{2}{\left[x-\frac{\sin (2x)}{2}\right]}_0^{\pi }=\frac{1}{2}\left[(\pi -0)-(0-0)\right]=\frac{\pi }{2}$.

   Key takeaway

   Use the half-angle identity $\sin^2 x = (1-\cos 2x)/2$ to integrate even powers of sine; $\sin(2\pi) = \sin(0) = 0$ eliminates the cosine term.

10. Question 10 · Hard

    Evaluate $\int x^2{e}^{x}dx$.

    • A
      $e^x(x^2-2x+2)+C$Correct
    • B
      $e^x(x^2+2x+2)+C$

      Why not B: Sign error in the recursive application: after two integrations by parts the $2x$ term should be $-2x$.
    • C
      $e^x(x^2-2x)+C$

      Why not C: Performs two rounds of IBP but misses the final $+2e^x$ term from $\int 2e^{x}dx=2e^x$.
    • D
      $x^2{e}^x-e^x+C$

      Why not D: Applies IBP once correctly but then fails to apply IBP a second time to $\int 2x{e}^{x}dx$, replacing it with $-e^x$ erroneously.
    Explanation

    IBP twice. First: $u=x^2$, $dv=e^{x}dx$: $x^2{e}^x-\int 2x{e}^{x}dx$. Second IBP on $\int 2x{e}^{x}dx$: $u=2x$, $dv=e^{x}dx$: $2x{e}^x-\int 2e^{x}dx=2x{e}^x-2e^x$. Full: $x^2{e}^x-(2x{e}^x-2e^x)+C=e^x(x^2-2x+2)+C$.

    Key takeaway

    For $\int x^n e^x\,dx$, apply integration by parts $n$ times; the pattern is $e^x(x^n - nx^{n-1} + n(n-1)x^{n-2} - \cdots)$.

11. Question 11 · Hard

    If $G(x)={\int }_1^{x^3}\ln tdt$, find $G^{\prime }(x)$.

    • A
      $\ln (x^3)$

      Why not A: Applies FTC Part 1 without the chain-rule factor from the upper limit $x^3$.
    • B
      $3x^2\ln (x^3)$Correct
    • C
      $\frac{3x^2}{x^3}$

      Why not C: Differentiates $\ln (x^3)$ rather than evaluating it; computes $d/dx[\ln (x^3)]\cdot \text{something}$.
    • D
      $3\ln x$

      Why not D: Uses $\ln (x^3)=3\ln x$ but then returns $3\ln x$ without the $3x^2$ chain factor.
    Explanation

    FTC Part 1 with chain rule: $G^{\prime }(x)=\ln (x^3)\cdot \frac{d}{dx}[x^3]=\ln (x^3)\cdot 3x^2=3x^2\ln (x^3)$. Equivalently $=9x^2\ln x$.

    Key takeaway

    When the upper limit is $g(x)$: $\frac{d}{dx}\int_a^{g(x)}f(t)\,dt = f(g(x))\cdot g'(x)$ by FTC + chain rule.

12. Question 12 · Hard

    Evaluate $\int \frac{x}{\sqrt{4-x^2}}dx$.

    • A
      $\arcsin \left(\frac{x}{2}\right)+C$

      Why not A: Applies the $\int 1/\sqrt{a^2-x^2}dx=\arcsin (x/a)$ formula but ignores the extra $x$ factor in the numerator.
    • B
      $-\sqrt{4-x^2}+C$Correct
    • C
      $\sqrt{4-x^2}+C$

      Why not C: Correct form but wrong sign; differentiating $+\sqrt{4-x^2}$ gives $-x/\sqrt{4-x^2}$, not $+x/\sqrt{4-x^2}$.
    • D
      $-\frac{1}{2}\sqrt{4-x^2}+C$

      Why not D: Introduces a spurious factor of $1/2$ from confusing with the $\int 1/\sqrt{u}$ pattern that requires a $1/2$ scaling.
    Explanation

    $u=4-x^2$, $du=-2xdx$, so $xdx=-\frac{1}{2}du$. $\int \frac{x}{\sqrt{4-x^2}}dx=\int \frac{-du/2}{\sqrt{u}}=-\frac{1}{2}\int u^{-1/2}du=-\frac{1}{2}\cdot 2\sqrt{u}+C=-\sqrt{4-x^2}+C$.

    Key takeaway

    Substitution $u = a^2 - x^2$ handles $\int x/\sqrt{a^2-x^2}\,dx$; the result is $-\sqrt{a^2-x^2} + C$, not an arcsin.