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Definite integrals practice problems

Evaluate with the Fundamental Theorem of Calculus: F(b) − F(a). 70 problems with worked solutions; in 70 of them every equation is proved by a computer algebra system.

20 worked problems

Try each one before opening the solution. Every line is labeled with what checked it. How we verify

Problem 4.1

medium Video
Evaluate \( \displaystyle \int_{-1}^{2} x + e^{2 x} \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(\frac{x^{2}}{2} + \frac{e^{2 x}}{2}\right) = x + e^{2 x} \]
    F(x) = x**2/2 + exp(2*x)/2 is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. \frac{x^{2}}{2} + \frac{e^{2 x}}{2} \right|_{\substack{ x=-1 }} + \left. \frac{x^{2}}{2} + \frac{e^{2 x}}{2} \right|_{\substack{ x=2 }} = \frac{-1 + \left(3 + e^{4}\right) e^{2}}{2 e^{2}} \]
    F(b) − F(a).✓ Proved
Answer \( \frac{-1 + \left(3 + e^{4}\right) e^{2}}{2 e^{2}} \)
The graph of f(x) = x + e^(2x) from x = -1 to x = 2. Rectangles under the curve get thinner until they fill the area, -e^(-2)/2 + 3/2 + e^(4)/2. Area below the x-axis counts as negative.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.7

medium Video
Evaluate \( \displaystyle \int_{-1}^{3} x + e^{x} \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(\frac{x^{2}}{2} + e^{x}\right) = x + e^{x} \]
    F(x) = x**2/2 + exp(x) is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. \frac{x^{2}}{2} + e^{x} \right|_{\substack{ x=-1 }} + \left. \frac{x^{2}}{2} + e^{x} \right|_{\substack{ x=3 }} = - \frac{1}{e} + 4 + e^{3} \]
    F(b) − F(a).✓ Proved
Answer \( - \frac{1}{e} + 4 + e^{3} \)
The graph of f(x) = x + e^(x) from x = -1 to x = 3. Rectangles under the curve get thinner until they fill the area, -e^(-1) + 4 + e^(3). Area below the x-axis counts as negative.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.10

medium Video
Evaluate \( \displaystyle \int_{0}^{\frac{\pi}{3}} 3 \sin{\left(x \right)} + 2 \cos{\left(x \right)} \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(2 \sin{\left(x \right)} - 3 \cos{\left(x \right)}\right) = 3 \sin{\left(x \right)} + 2 \cos{\left(x \right)} \]
    F(x) = 2*sin(x) - 3*cos(x) is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. 2 \sin{\left(x \right)} - 3 \cos{\left(x \right)} \right|_{\substack{ x=0 }} + \left. 2 \sin{\left(x \right)} - 3 \cos{\left(x \right)} \right|_{\substack{ x=\frac{\pi}{3} }} = \frac{3}{2} + \sqrt{3} \]
    F(b) − F(a).✓ Proved
Answer \( \frac{3}{2} + \sqrt{3} \)
The graph of f(x) = 3sin(x) + 2cos(x) from x = 0 to x = pi/3. Rectangles under the curve get thinner until they fill the area, 3/2 + sqrt(3).

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.47

medium Video
Evaluate \( \displaystyle \int_{1}^{2} x + \frac{3}{x} \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(\frac{x^{2}}{2} + 3 \ln{\left(x \right)}\right) = x + \frac{3}{x} \]
    F(x) = x**2/2 + 3*log(x) is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. \frac{x^{2}}{2} + 3 \ln{\left(x \right)} \right|_{\substack{ x=1 }} + \left. \frac{x^{2}}{2} + 3 \ln{\left(x \right)} \right|_{\substack{ x=2 }} = \frac{3}{2} + \ln{\left(8 \right)} \]
    F(b) − F(a).✓ Proved
Answer \( \frac{3}{2} + \ln{\left(8 \right)} \)
The graph of f(x) = x + 3/x from x = 1 to x = 2. Rectangles under the curve get thinner until they fill the area, 3/2 + 3log(2).

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.49

medium Video
Evaluate \( \displaystyle \int_{1}^{2} \sqrt{x} - 1 \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(\frac{2 x^{\frac{3}{2}}}{3} - x\right) = \sqrt{x} - 1 \]
    F(x) = 2*x**(3/2)/3 - x is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. - x + \frac{2 x^{\frac{3}{2}}}{3} \right|_{\substack{ x=1 }} + \left. - x + \frac{2 x^{\frac{3}{2}}}{3} \right|_{\substack{ x=2 }} = - \frac{5}{3} + \frac{4 \sqrt{2}}{3} \]
    F(b) − F(a).✓ Proved
Answer \( - \frac{5}{3} + \frac{4 \sqrt{2}}{3} \)
The graph of f(x) = sqrt(x) - 1 from x = 1 to x = 2. Rectangles under the curve get thinner until they fill the area, -5/3 + 4sqrt(2)/3.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.50

medium Video
Evaluate \( \displaystyle \int_{1}^{e} 2 \sqrt{x} - 2 \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(\frac{4 x^{\frac{3}{2}}}{3} - 2 x\right) = 2 \sqrt{x} - 2 \]
    F(x) = 4*x**(3/2)/3 - 2*x is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. - 2 x + \frac{4 x^{\frac{3}{2}}}{3} \right|_{\substack{ x=1 }} + \left. - 2 x + \frac{4 x^{\frac{3}{2}}}{3} \right|_{\substack{ x=e }} = - 2 e + \frac{2}{3} + \frac{4 e^{\frac{3}{2}}}{3} \]
    F(b) − F(a).✓ Proved
Answer \( - 2 e + \frac{2}{3} + \frac{4 e^{\frac{3}{2}}}{3} \)
The graph of f(x) = 2sqrt(x) - 2 from x = 1 to x = E. Rectangles under the curve get thinner until they fill the area, -2E + 2/3 + 4e^(3/2)/3.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.58

medium Video
Evaluate \( \displaystyle \int_{1}^{2} 2 \sqrt{x} - 1 \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(\frac{4 x^{\frac{3}{2}}}{3} - x\right) = 2 \sqrt{x} - 1 \]
    F(x) = 4*x**(3/2)/3 - x is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. - x + \frac{4 x^{\frac{3}{2}}}{3} \right|_{\substack{ x=1 }} + \left. - x + \frac{4 x^{\frac{3}{2}}}{3} \right|_{\substack{ x=2 }} = - \frac{7}{3} + \frac{8 \sqrt{2}}{3} \]
    F(b) − F(a).✓ Proved
Answer \( - \frac{7}{3} + \frac{8 \sqrt{2}}{3} \)
The graph of f(x) = 2sqrt(x) - 1 from x = 1 to x = 2. Rectangles under the curve get thinner until they fill the area, -7/3 + 8sqrt(2)/3.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.91

medium Video
Evaluate \( \displaystyle \int_{1}^{2} 3 \sqrt{x} + 1 \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(2 x^{\frac{3}{2}} + x\right) = 3 \sqrt{x} + 1 \]
    F(x) = 2*x**(3/2) + x is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. x + 2 x^{\frac{3}{2}} \right|_{\substack{ x=1 }} + \left. x + 2 x^{\frac{3}{2}} \right|_{\substack{ x=2 }} = -1 + 4 \sqrt{2} \]
    F(b) − F(a).✓ Proved
Answer \( -1 + 4 \sqrt{2} \)
The graph of f(x) = 3sqrt(x) + 1 from x = 1 to x = 2. Rectangles under the curve get thinner until they fill the area, -1 + 4sqrt(2).

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.92

medium Video
Evaluate \( \displaystyle \int_{0}^{\frac{\pi}{3}} 2 \sin{\left(x \right)} - 2 \cos{\left(x \right)} \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(- 2 \sin{\left(x \right)} - 2 \cos{\left(x \right)}\right) = 2 \sin{\left(x \right)} - 2 \cos{\left(x \right)} \]
    F(x) = -2*sin(x) - 2*cos(x) is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. - 2 \sin{\left(x \right)} - 2 \cos{\left(x \right)} \right|_{\substack{ x=0 }} + \left. - 2 \sin{\left(x \right)} - 2 \cos{\left(x \right)} \right|_{\substack{ x=\frac{\pi}{3} }} = 1 - \sqrt{3} \]
    F(b) − F(a).✓ Proved
Answer \( 1 - \sqrt{3} \)
The graph of f(x) = 2sin(x) - 2cos(x) from x = 0 to x = pi/3. Rectangles under the curve get thinner until they fill the area, 1 - sqrt(3). Area below the x-axis counts as negative.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.106

medium Video
Evaluate \( \displaystyle \int_{1}^{e} x + \frac{2}{x} \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(\frac{x^{2}}{2} + 2 \ln{\left(x \right)}\right) = x + \frac{2}{x} \]
    F(x) = x**2/2 + 2*log(x) is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. \frac{x^{2}}{2} + 2 \ln{\left(x \right)} \right|_{\substack{ x=1 }} + \left. \frac{x^{2}}{2} + 2 \ln{\left(x \right)} \right|_{\substack{ x=e }} = \frac{3}{2} + \frac{e^{2}}{2} \]
    F(b) − F(a).✓ Proved
Answer \( \frac{3}{2} + \frac{e^{2}}{2} \)
The graph of f(x) = x + 2/x from x = 1 to x = E. Rectangles under the curve get thinner until they fill the area, 3/2 + e^(2)/2.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.108

medium Video
Evaluate \( \displaystyle \int_{2}^{3} x + e^{x} \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(\frac{x^{2}}{2} + e^{x}\right) = x + e^{x} \]
    F(x) = x**2/2 + exp(x) is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. \frac{x^{2}}{2} + e^{x} \right|_{\substack{ x=2 }} + \left. \frac{x^{2}}{2} + e^{x} \right|_{\substack{ x=3 }} = - e^{2} + \frac{5}{2} + e^{3} \]
    F(b) − F(a).✓ Proved
Answer \( - e^{2} + \frac{5}{2} + e^{3} \)
The graph of f(x) = x + e^(x) from x = 2 to x = 3. Rectangles under the curve get thinner until they fill the area, -e^(2) + 5/2 + e^(3).

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.201

medium Video
Evaluate \( \displaystyle \int_{1}^{e} 4 \sqrt{x} + 2 \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(\frac{8 x^{\frac{3}{2}}}{3} + 2 x\right) = 4 \sqrt{x} + 2 \]
    F(x) = 8*x**(3/2)/3 + 2*x is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. 2 x + \frac{8 x^{\frac{3}{2}}}{3} \right|_{\substack{ x=1 }} + \left. 2 x + \frac{8 x^{\frac{3}{2}}}{3} \right|_{\substack{ x=e }} = - \frac{14}{3} + 2 e + \frac{8 e^{\frac{3}{2}}}{3} \]
    F(b) − F(a).✓ Proved
Answer \( - \frac{14}{3} + 2 e + \frac{8 e^{\frac{3}{2}}}{3} \)
The graph of f(x) = 4sqrt(x) + 2 from x = 1 to x = E. Rectangles under the curve get thinner until they fill the area, -14/3 + 2E + 8e^(3/2)/3.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.2

easy Video
Evaluate \( \displaystyle \int_{2}^{3} 2 - 2 x \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(- x^{2} + 2 x\right) = 2 - 2 x \]
    F(x) = -x**2 + 2*x is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. - x^{2} + 2 x \right|_{\substack{ x=2 }} + \left. - x^{2} + 2 x \right|_{\substack{ x=3 }} = -3 \]
    F(b) − F(a).✓ Proved
Answer \( -3 \)
The graph of f(x) = 2 - 2x from x = 2 to x = 3. Rectangles under the curve get thinner until they fill the area, -3. Area below the x-axis counts as negative.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.3

easy Video
Evaluate \( \displaystyle \int_{1}^{3} 4 x - 2 \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(2 x^{2} - 2 x\right) = 4 x - 2 \]
    F(x) = 2*x**2 - 2*x is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. 2 x^{2} - 2 x \right|_{\substack{ x=1 }} + \left. 2 x^{2} - 2 x \right|_{\substack{ x=3 }} = 12 \]
    F(b) − F(a).✓ Proved
Answer \( 12 \)
The graph of f(x) = 4x - 2 from x = 1 to x = 3. Rectangles under the curve get thinner until they fill the area, 12.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.4

easy Video
Evaluate \( \displaystyle \int_{-2}^{1} x \left(x^{2} + 1\right)^{3} \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(\frac{x^{8}}{8} + \frac{x^{6}}{2} + \frac{3 x^{4}}{4} + \frac{x^{2}}{2}\right) = x \left(x^{2} + 1\right)^{3} \]
    F(x) = x**8/8 + x**6/2 + 3*x**4/4 + x**2/2 is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. \frac{x^{8}}{8} + \frac{x^{6}}{2} + \frac{3 x^{4}}{4} + \frac{x^{2}}{2} \right|_{\substack{ x=-2 }} + \left. \frac{x^{8}}{8} + \frac{x^{6}}{2} + \frac{3 x^{4}}{4} + \frac{x^{2}}{2} \right|_{\substack{ x=1 }} = - \frac{609}{8} \]
    F(b) − F(a).✓ Proved
Answer \( - \frac{609}{8} \)
The graph of f(x) = x·(x² + 1)³ from x = -2 to x = 1. Rectangles under the curve get thinner until they fill the area, -609/8. Area below the x-axis counts as negative.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.5

easy Video
Evaluate \( \displaystyle \int_{-2}^{3} 4 x^{3} + 3 x^{2} - 2 \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(x^{4} + x^{3} - 2 x\right) = 4 x^{3} + 3 x^{2} - 2 \]
    F(x) = x**4 + x**3 - 2*x is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. x^{4} + x^{3} - 2 x \right|_{\substack{ x=-2 }} + \left. x^{4} + x^{3} - 2 x \right|_{\substack{ x=3 }} = 90 \]
    F(b) − F(a).✓ Proved
Answer \( 90 \)
The graph of f(x) = 4x³ + 3x² - 2 from x = -2 to x = 3. Rectangles under the curve get thinner until they fill the area, 90. Area below the x-axis counts as negative.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.8

easy Video
Evaluate \( \displaystyle \int_{0}^{3} - 3 x^{3} - 4 x^{2} - 3 x + 1 \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(- \frac{3 x^{4}}{4} - \frac{4 x^{3}}{3} - \frac{3 x^{2}}{2} + x\right) = - 3 x^{3} - 4 x^{2} - 3 x + 1 \]
    F(x) = -3*x**4/4 - 4*x**3/3 - 3*x**2/2 + x is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. - \frac{3 x^{4}}{4} - \frac{4 x^{3}}{3} - \frac{3 x^{2}}{2} + x \right|_{\substack{ x=0 }} + \left. - \frac{3 x^{4}}{4} - \frac{4 x^{3}}{3} - \frac{3 x^{2}}{2} + x \right|_{\substack{ x=3 }} = - \frac{429}{4} \]
    F(b) − F(a).✓ Proved
Answer \( - \frac{429}{4} \)
The graph of f(x) = -3x³ - 4x² - 3x + 1 from x = 0 to x = 3. Rectangles under the curve get thinner until they fill the area, -429/4. Area below the x-axis counts as negative.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.13

easy Video
Evaluate \( \displaystyle \int_{1}^{2} x \left(x^{2} + 1\right)^{3} \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(\frac{x^{8}}{8} + \frac{x^{6}}{2} + \frac{3 x^{4}}{4} + \frac{x^{2}}{2}\right) = x \left(x^{2} + 1\right)^{3} \]
    F(x) = x**8/8 + x**6/2 + 3*x**4/4 + x**2/2 is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. \frac{x^{8}}{8} + \frac{x^{6}}{2} + \frac{3 x^{4}}{4} + \frac{x^{2}}{2} \right|_{\substack{ x=1 }} + \left. \frac{x^{8}}{8} + \frac{x^{6}}{2} + \frac{3 x^{4}}{4} + \frac{x^{2}}{2} \right|_{\substack{ x=2 }} = \frac{609}{8} \]
    F(b) − F(a).✓ Proved
Answer \( \frac{609}{8} \)
The graph of f(x) = x·(x² + 1)³ from x = 1 to x = 2. Rectangles under the curve get thinner until they fill the area, 609/8.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.15

easy Video
Evaluate \( \displaystyle \int_{1}^{3} 2 x^{2} - 2 x + 3 \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(\frac{2 x^{3}}{3} - x^{2} + 3 x\right) = 2 x^{2} - 2 x + 3 \]
    F(x) = 2*x**3/3 - x**2 + 3*x is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. \frac{2 x^{3}}{3} - x^{2} + 3 x \right|_{\substack{ x=1 }} + \left. \frac{2 x^{3}}{3} - x^{2} + 3 x \right|_{\substack{ x=3 }} = \frac{46}{3} \]
    F(b) − F(a).✓ Proved
Answer \( \frac{46}{3} \)
The graph of f(x) = 2x² - 2x + 3 from x = 1 to x = 3. Rectangles under the curve get thinner until they fill the area, 46/3.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.53

easy Video
Evaluate \( \displaystyle \int_{0}^{1} 3 x^{2} - 3 x - 3 \, dx \).
  1. By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
    Reviewed
  2. \[ \frac{d}{d x} \left(x^{3} - \frac{3 x^{2}}{2} - 3 x\right) = 3 x^{2} - 3 x - 3 \]
    F(x) = x**3 - 3*x**2/2 - 3*x is an antiderivative: its derivative is the integrand.✓ Proved
  3. \[ - \left. x^{3} - \frac{3 x^{2}}{2} - 3 x \right|_{\substack{ x=0 }} + \left. x^{3} - \frac{3 x^{2}}{2} - 3 x \right|_{\substack{ x=1 }} = - \frac{7}{2} \]
    F(b) − F(a).✓ Proved
Answer \( - \frac{7}{2} \)
The graph of f(x) = 3x² - 3x - 3 from x = 0 to x = 1. Rectangles under the curve get thinner until they fill the area, -7/2. Area below the x-axis counts as negative.

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

More definite integrals practice

Evaluate \( \displaystyle \int_{1}^{9} 3 \sqrt{x} - 2 \, dx \).
Problem 4.6easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{\frac{\pi}{2}} 2 \sin{\left(x \right)} + 2 \cos{\left(x \right)} \, dx \).
Problem 4.11easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{9} 4 \sqrt{x} + 2 \, dx \).
Problem 4.46easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{3} x \left(x^{2} + 1\right)^{2} \, dx \).
Problem 4.48easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{4} 2 \sqrt{x} + 2 \, dx \).
Problem 4.51easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{2} x + e^{x} \, dx \).
Problem 4.52easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{\frac{\pi}{2}} 3 \sin{\left(x \right)} + \cos{\left(x \right)} \, dx \).
Problem 4.54easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{9} x + \frac{3}{x} \, dx \).
Problem 4.55easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-2}^{3} x \left(x^{2} + 1\right)^{2} \, dx \).
Problem 4.56easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{\pi} 3 \sin{\left(x \right)} - 3 \cos{\left(x \right)} \, dx \).
Problem 4.57easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{2} x \left(x^{2} + 1\right)^{2} \, dx \).
Problem 4.60easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-1}^{2} x \left(x^{2} + 1\right)^{3} \, dx \).
Problem 4.88easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-1}^{1} x \left(x^{2} + 1\right)^{3} \, dx \).
Problem 4.93easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{2}^{3} x \left(x^{2} + 1\right)^{3} \, dx \).
Problem 4.94easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{4} 3 \sqrt{x} - 1 \, dx \).
Problem 4.95easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{\frac{\pi}{2}} 3 \sin{\left(x \right)} - 3 \cos{\left(x \right)} \, dx \).
Problem 4.105easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-2}^{-1} x \left(x^{2} + 1\right)^{2} \, dx \).
Problem 4.107easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{\pi} 2 \sin{\left(x \right)} - 3 \cos{\left(x \right)} \, dx \).
Problem 4.110easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-2}^{3} x \left(x^{2} + 1\right)^{3} \, dx \).
Problem 4.122easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{1} 3 x^{2} - 3 x + 2 \, dx \).
Problem 4.123easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-1}^{3} - 2 x^{2} - 4 x + 1 \, dx \).
Problem 4.124easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{4} \sqrt{x} + 1 \, dx \).
Problem 4.126easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-2}^{2} 2 x^{3} + 2 x^{2} + 3 x + 4 \, dx \).
Problem 4.127easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-1}^{0} x \left(x^{2} + 1\right)^{2} \, dx \).
Problem 4.128easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{-2}^{2} x + e^{x} \, dx \).
Problem 4.200easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-2}^{2} 3 x - 3 \, dx \).
Problem 4.202easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{\frac{\pi}{2}} \sin{\left(x \right)} + 3 \cos{\left(x \right)} \, dx \).
Problem 4.203easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-2}^{3} - 4 x^{3} + x^{2} + 3 x - 1 \, dx \).
Problem 4.204easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{9} x + \frac{1}{x} \, dx \).
Problem 4.205easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{2} - 2 x^{2} - 2 x + 4 \, dx \).
Problem 4.206easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{2} 1 - 2 x \, dx \).
Problem 4.236easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-2}^{2} x + e^{2 x} \, dx \).
Problem 4.237easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{3} 3 - 4 x \, dx \).
Problem 4.239easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{\frac{\pi}{2}} \sin{\left(x \right)} - 3 \cos{\left(x \right)} \, dx \).
Problem 4.240easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{2}^{3} 1 - 3 x \, dx \).
Problem 4.241easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{4} 4 \sqrt{x} - 3 \, dx \).
Problem 4.242easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{\pi} 3 \sin{\left(x \right)} + 3 \cos{\left(x \right)} \, dx \).
Problem 4.243easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-2}^{3} x + e^{2 x} \, dx \).
Problem 4.9medium Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{4} x + \frac{5}{x} \, dx \).
Problem 4.12medium Video✓ Every equation proved
Evaluate \( \displaystyle \int_{-1}^{0} x + e^{x} \, dx \).
Problem 4.14medium Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{2} 3 \sqrt{x} - 2 \, dx \).
Problem 4.59medium Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{\frac{\pi}{3}} 2 \sin{\left(x \right)} + 2 \cos{\left(x \right)} \, dx \).
Problem 4.89medium Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-1}^{3} x + e^{2 x} \, dx \).
Problem 4.90medium Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{e} 3 \sqrt{x} - 2 \, dx \).
Problem 4.104medium Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{3} x + e^{x} \, dx \).
Problem 4.109medium Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{e} 4 \sqrt{x} - 2 \, dx \).
Problem 4.111medium Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{e} x + \frac{5}{x} \, dx \).
Problem 4.121medium Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{4} x + \frac{3}{x} \, dx \).
Problem 4.125medium Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-2}^{0} x + e^{2 x} \, dx \).
Problem 4.199medium Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{9} x + \frac{2}{x} \, dx \).
Problem 4.238medium Video✓ Nihil obstat