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Antiderivatives

17 practice questions with full worked solutions. Free, no account needed.

Problems & worked solutions

Problem #0122 International

Problem 1 — Antiderivatives

∫5 dx\displaystyle\int 5 \, dx equals:

Show answer & worked solution
  1. A. 00
  2. B. 55
  3. C. 5x+C5x + C✓ correct
  4. D. x25+C\dfrac{x^2}{5} + C

∫c dx=cx+C\int c \, dx = cx + C. So ∫5 dx=5x+C\int 5 \, dx = 5x + C.

Problem #0123 International

Problem 2 — Antiderivatives

∫ex dx\displaystyle\int e^x \, dx equals:

Show answer & worked solution
  1. A. ex+1+Ce^{x+1} + C
  2. B. xex+Cx e^x + C
  3. C. ex+Ce^x + C✓ correct
  4. D. ln⁡x+C\ln x + C

The exponential is its own antiderivative: ∫ex dx=ex+C\int e^x \, dx = e^x + C.

Problem #0121 International

Problem 3 — Antiderivatives

∫x3 dx\displaystyle\int x^3 \, dx equals:

Show answer & worked solution
  1. A. 3x2+C3x^2 + C
  2. B. x43+C\dfrac{x^4}{3} + C
  3. C. x44+C\dfrac{x^4}{4} + C✓ correct
  4. D. 4x4+C4x^4 + C

∫x3 dx=x44+C\int x^3 \, dx = \dfrac{x^4}{4} + C.

Problem #0001 International

Problem 4 — Definite Integrals

Evaluate ∫0π/2sin⁡3(x)cos⁡(x) dx\int_0^{\pi/2} \sin^3(x)\cos(x)\,dx.

Show answer & worked solution
  1. A. 18\dfrac{1}{8}
  2. B. 14\dfrac{1}{4}✓ correct
  3. C. 38\dfrac{3}{8}
  4. D. 12\dfrac{1}{2}

Let u=sin⁡xu = \sin x, du=cos⁡x dxdu = \cos x\,dx. The integral becomes ∫01u3 du=[u44]01=14\int_0^1 u^3\,du = \left[\dfrac{u^4}{4}\right]_0^1 = \dfrac{1}{4}.

Problem #0124 International

Problem 5 — Antiderivatives

∫1x dx\displaystyle\int \dfrac{1}{x} \, dx (for x>0x > 0) equals:

Show answer & worked solution
  1. A. 1x2+C\dfrac{1}{x^2} + C
  2. B. −1x2+C-\dfrac{1}{x^2} + C
  3. C. ln⁡x+C\ln x + C✓ correct
  4. D. x00+C\dfrac{x^0}{0} + C

∫dxx=ln⁡∣x∣+C\int \dfrac{dx}{x} = \ln|x| + C. For x>0x > 0: ln⁡x+C\ln x + C.

Problem #0126 International

Problem 6 — Antiderivatives

∫xex dx\displaystyle\int x e^x \, dx equals:

Show answer & worked solution
  1. A. x22ex+C\dfrac{x^2}{2}e^x + C
  2. B. ex+Ce^x + C
  3. C. (x−1)ex+C(x - 1)e^x + C✓ correct
  4. D. xex+Cx e^x + C

∫xex dx=xex−∫ex dx=xex−ex+C=(x−1)ex+C\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C = (x - 1) e^x + C.

Problem #0128 International

Problem 7 — Antiderivatives

∫1x2+1 dx\displaystyle\int \dfrac{1}{x^2 + 1} \, dx equals:

Show answer & worked solution
  1. A. ln⁡(x2+1)+C\ln(x^2 + 1) + C
  2. B. 2xx2+1+C\dfrac{2x}{x^2 + 1} + C
  3. C. arctan⁡x+C\arctan x + C✓ correct
  4. D. arcsin⁡x+C\arcsin x + C

A standard antiderivative: ∫dx1+x2=arctan⁡x+C\int \dfrac{dx}{1 + x^2} = \arctan x + C.

Problem #0127 International

Problem 8 — Antiderivatives

∫2x⋅cos⁡(x2) dx\displaystyle\int 2x \cdot \cos(x^2) \, dx equals:

Show answer & worked solution
  1. A. sin⁡(2x)+C\sin(2x) + C
  2. B. 2sin⁡(x2)+C2\sin(x^2) + C
  3. C. sin⁡(x2)+C\sin(x^2) + C✓ correct
  4. D. cos⁡(x2)+C\cos(x^2) + C

u=x2⇒du=2x dxu = x^2 \Rightarrow du = 2x \, dx. So ∫cos⁡u du=sin⁡u+C=sin⁡(x2)+C\int \cos u \, du = \sin u + C = \sin(x^2) + C.

Problem #0125 International

Problem 9 — Antiderivatives

∫cos⁡x dx\displaystyle\int \cos x \, dx equals:

Show answer & worked solution
  1. A. −cos⁡x+C-\cos x + C
  2. B. −sin⁡x+C-\sin x + C
  3. C. sin⁡x+C\sin x + C✓ correct
  4. D. tan⁡x+C\tan x + C

∫cos⁡x dx=sin⁡x+C\int \cos x \, dx = \sin x + C.

Problem #0014 International

Problem 10 — Integration by Parts

Compute ∫01xex dx\displaystyle\int_0^1 x e^x\, dx.

Show answer & worked solution
  1. A. 00
  2. B. 11✓ correct
  3. C. e−1e - 1
  4. D. e+1e + 1

∫xex dx=xex−∫ex dx=(x−1)ex+C\int x e^x\,dx = x e^x - \int e^x\,dx = (x-1)e^x + C. Evaluating from 00 to 11: (0)⋅e−(−1)⋅1=0−(−1)=1(0)\cdot e - (-1)\cdot 1 = 0 - (-1) = 1.

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