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Definite integrals from area formulas practice problems

Evaluate a definite integral as a signed area: triangles, rectangles, trapezoids and semicircles. 60 problems with worked solutions; in 60 of them every equation is proved by a computer algebra system.

14 worked problems

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

Problem 4.1062

easy Video
Evaluate \( \displaystyle \int_{1}^{2} 2 x + 2 \, dx \) using a formula from geometry.
  1. The region between the graph and the x-axis is a trapezoid.
    Reviewed
  2. \[ \frac{\left(-1 + 2\right) \left(4 + 6\right)}{2} = 5 \]
    Its (signed) area.✓ Proved
  3. \[ \int\limits_{1}^{2} \left(2 x + 2\right)\, dx = 5 \]
    Integrating directly gives the same.✓ Proved
Answer \( 5 \)
The graph of f(x) = 2x + 2 from x = 1 to x = 2. Rectangles under the curve get thinner until they fill the area, 5.

✓ 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 4.1070

easy Video
Evaluate \( \displaystyle \int_{0}^{1} \sqrt{1 - x^{2}} \, dx \) using a formula from geometry.
  1. The region between the graph and the x-axis is a quarter circle of radius 1.
    Reviewed
  2. \[ \frac{1^{2} \pi}{4} = \frac{\pi}{4} \]
    Its (signed) area.✓ Proved
  3. \[ \int\limits_{0}^{1} \sqrt{1 - x^{2}}\, dx = \frac{\pi}{4} \]
    Integrating directly gives the same.✓ Proved
Answer \( \frac{\pi}{4} \)
The graph of f(x) = sqrt(1 - x²) from x = 0 to x = 1. Rectangles under the curve get thinner until they fill the area, pi/4.

✓ 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 4.1075

easy Video
Evaluate \( \displaystyle \int_{-3}^{2} \left|{x + 2}\right| \, dx \) using a formula from geometry.
  1. The region between the graph and the x-axis is two triangles meeting at x = -2.
    Reviewed
  2. \[ \frac{1^{2}}{2} + \frac{4^{2}}{2} = \frac{17}{2} \]
    Its (signed) area.✓ Proved
  3. \[ \int\limits_{-3}^{2} \left|{x + 2}\right|\, dx = \frac{17}{2} \]
    Integrating directly gives the same.✓ Proved
Answer \( \frac{17}{2} \)
The graph of f(x) = Abs(x + 2) from x = -3 to x = 2. Rectangles under the curve get thinner until they fill the area, 17/2.

✓ 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 4.1141

easy Video
Evaluate \( \displaystyle \int_{-1}^{1} \sqrt{1 - x^{2}} \, dx \) using a formula from geometry.
  1. The region between the graph and the x-axis is a semicircle of radius 1.
    Reviewed
  2. \[ \frac{1^{2} \pi}{2} = \frac{\pi}{2} \]
    Its (signed) area.✓ Proved
  3. \[ \int\limits_{-1}^{1} \sqrt{1 - x^{2}}\, dx = \frac{\pi}{2} \]
    Integrating directly gives the same.✓ Proved
Answer \( \frac{\pi}{2} \)
The graph of f(x) = sqrt(1 - x²) from x = -1 to x = 1. Rectangles under the curve get thinner until they fill the area, pi/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 4.1142

easy Video
Evaluate \( \displaystyle \int_{1}^{5} \left|{x - 2}\right| \, dx \) using a formula from geometry.
  1. The region between the graph and the x-axis is two triangles meeting at x = 2.
    Reviewed
  2. \[ \frac{1^{2}}{2} + \frac{3^{2}}{2} = 5 \]
    Its (signed) area.✓ Proved
  3. \[ \int\limits_{1}^{5} \left|{x - 2}\right|\, dx = 5 \]
    Integrating directly gives the same.✓ Proved
Answer \( 5 \)
The graph of f(x) = Abs(x - 2) from x = 1 to x = 5. Rectangles under the curve get thinner until they fill the area, 5.

✓ 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 4.1143

easy Video
Evaluate \( \displaystyle \int_{1}^{5} 2 - 2 x \, dx \) using a formula from geometry.
  1. The region between the graph and the x-axis is a trapezoid (a triangle, since one side has height 0), below the axis.
    Reviewed
  2. \[ \frac{\left(-8 + 0\right) \left(-1 + 5\right)}{2} = -16 \]
    Its (signed) area.✓ Proved
  3. \[ \int\limits_{1}^{5} \left(2 - 2 x\right)\, dx = -16 \]
    Integrating directly gives the same.✓ Proved
Answer \( -16 \)
The graph of f(x) = 2 - 2x from x = 1 to x = 5. Rectangles under the curve get thinner until they fill the area, -16. 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 4.1145

easy Video
Evaluate \( \displaystyle \int_{0}^{2} x + 1 \, dx \) using a formula from geometry.
  1. The region between the graph and the x-axis is a trapezoid.
    Reviewed
  2. \[ \frac{\left(0 + 2\right) \left(1 + 3\right)}{2} = 4 \]
    Its (signed) area.✓ Proved
  3. \[ \int\limits_{0}^{2} \left(x + 1\right)\, dx = 4 \]
    Integrating directly gives the same.✓ Proved
Answer \( 4 \)
The graph of f(x) = x + 1 from x = 0 to x = 2. Rectangles under the curve get thinner until they fill the area, 4.

✓ 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 4.1147

easy Video
Evaluate \( \displaystyle \int_{-2}^{6} \left|{x - 2}\right| \, dx \) using a formula from geometry.
  1. The region between the graph and the x-axis is two triangles meeting at x = 2.
    Reviewed
  2. \[ \frac{4^{2}}{2} + \frac{4^{2}}{2} = 16 \]
    Its (signed) area.✓ Proved
  3. \[ \int\limits_{-2}^{6} \left|{x - 2}\right|\, dx = 16 \]
    Integrating directly gives the same.✓ Proved
Answer \( 16 \)
The graph of f(x) = Abs(x - 2) from x = -2 to x = 6. Rectangles under the curve get thinner until they fill the area, 16.

✓ 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 4.1190

easy Video
Evaluate \( \displaystyle \int_{1}^{4} 3 x \, dx \) using a formula from geometry.
  1. The region between the graph and the x-axis is a trapezoid.
    Reviewed
  2. \[ \frac{\left(-1 + 4\right) \left(3 + 12\right)}{2} = \frac{45}{2} \]
    Its (signed) area.✓ Proved
  3. \[ \int\limits_{1}^{4} 3 x\, dx = \frac{45}{2} \]
    Integrating directly gives the same.✓ Proved
Answer \( \frac{45}{2} \)
The graph of f(x) = 3x from x = 1 to x = 4. Rectangles under the curve get thinner until they fill the area, 45/2.

✓ 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 4.1191

easy Video
Evaluate \( \displaystyle \int_{-1}^{1} x + 5 \, dx \) using a formula from geometry.
  1. The region between the graph and the x-axis is a trapezoid.
    Reviewed
  2. \[ \frac{\left(1 + 1\right) \left(4 + 6\right)}{2} = 10 \]
    Its (signed) area.✓ Proved
  3. \[ \int\limits_{-1}^{1} \left(x + 5\right)\, dx = 10 \]
    Integrating directly gives the same.✓ Proved
Answer \( 10 \)
The graph of f(x) = x + 5 from x = -1 to x = 1. Rectangles under the curve get thinner until they fill the area, 10.

✓ 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 4.1192

easy Video
Evaluate \( \displaystyle \int_{-4}^{-1} \left|{x + 2}\right| \, dx \) using a formula from geometry.
  1. The region between the graph and the x-axis is two triangles meeting at x = -2.
    Reviewed
  2. \[ \frac{1^{2}}{2} + \frac{2^{2}}{2} = \frac{5}{2} \]
    Its (signed) area.✓ Proved
  3. \[ \int\limits_{-4}^{-1} \left|{x + 2}\right|\, dx = \frac{5}{2} \]
    Integrating directly gives the same.✓ Proved
Answer \( \frac{5}{2} \)
The graph of f(x) = Abs(x + 2) from x = -4 to x = -1. Rectangles under the curve get thinner until they fill the area, 5/2.

✓ 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 4.1195

easy Video
Evaluate \( \displaystyle \int_{-2}^{1} - x - 2 \, dx \) using a formula from geometry.
  1. The region between the graph and the x-axis is a trapezoid (a triangle, since one side has height 0), below the axis.
    Reviewed
  2. \[ \frac{\left(-3 + 0\right) \left(1 + 2\right)}{2} = - \frac{9}{2} \]
    Its (signed) area.✓ Proved
  3. \[ \int\limits_{-2}^{1} \left(- x - 2\right)\, dx = - \frac{9}{2} \]
    Integrating directly gives the same.✓ Proved
Answer \( - \frac{9}{2} \)
The graph of f(x) = -x - 2 from x = -2 to x = 1. Rectangles under the curve get thinner until they fill the area, -9/2. 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 4.1369

easy Video
Evaluate \( \displaystyle \int_{-1}^{4} 2 x + 2 \, dx \) using a formula from geometry.
  1. The region between the graph and the x-axis is a trapezoid (a triangle, since one side has height 0).
    Reviewed
  2. \[ \frac{\left(0 + 10\right) \left(1 + 4\right)}{2} = 25 \]
    Its (signed) area.✓ Proved
  3. \[ \int\limits_{-1}^{4} \left(2 x + 2\right)\, dx = 25 \]
    Integrating directly gives the same.✓ Proved
Answer \( 25 \)
The graph of f(x) = 2x + 2 from x = -1 to x = 4. Rectangles under the curve get thinner until they fill the area, 25.

✓ 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 4.1371

easy Video
Evaluate \( \displaystyle \int_{-1}^{0} 2 - x \, dx \) using a formula from geometry.
  1. The region between the graph and the x-axis is a trapezoid.
    Reviewed
  2. \[ \frac{\left(0 + 1\right) \left(2 + 3\right)}{2} = \frac{5}{2} \]
    Its (signed) area.✓ Proved
  3. \[ \int\limits_{-1}^{0} \left(2 - x\right)\, dx = \frac{5}{2} \]
    Integrating directly gives the same.✓ Proved
Answer \( \frac{5}{2} \)
The graph of f(x) = 2 - x from x = -1 to x = 0. Rectangles under the curve get thinner until they fill the area, 5/2.

✓ 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

More definite integrals from area formulas practice

Evaluate \( \displaystyle \int_{3}^{4} 1 - 3 x \, dx \) using a formula from geometry.
Problem 4.1063easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{-1}^{1} - x - 1 \, dx \) using a formula from geometry.
Problem 4.1064easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{1}^{5} - 3 x \, dx \) using a formula from geometry.
Problem 4.1065easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{-2}^{3} \left|{x}\right| \, dx \) using a formula from geometry.
Problem 4.1066easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{-4}^{1} \left|{x + 2}\right| \, dx \) using a formula from geometry.
Problem 4.1067easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{0}^{1} - 2 x - 1 \, dx \) using a formula from geometry.
Problem 4.1068easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{0}^{6} \left|{x - 2}\right| \, dx \) using a formula from geometry.
Problem 4.1069easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{-1}^{2} 3 x + 4 \, dx \) using a formula from geometry.
Problem 4.1071easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{2}^{4} 2 x + 4 \, dx \) using a formula from geometry.
Problem 4.1072easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{0}^{1} 2 x + 6 \, dx \) using a formula from geometry.
Problem 4.1073easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-4}^{1} \left|{x + 1}\right| \, dx \) using a formula from geometry.
Problem 4.1074easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{0}^{5} \sqrt{25 - x^{2}} \, dx \) using a formula from geometry.
Problem 4.1076easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{3} \sqrt{9 - x^{2}} \, dx \) using a formula from geometry.
Problem 4.1140easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{-2}^{3} \left|{x + 1}\right| \, dx \) using a formula from geometry.
Problem 4.1144easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{4}^{5} 1 - x \, dx \) using a formula from geometry.
Problem 4.1146easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{-2}^{2} \sqrt{4 - x^{2}} \, dx \) using a formula from geometry.
Problem 4.1148easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-1}^{4} 3 x + 4 \, dx \) using a formula from geometry.
Problem 4.1149easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{2}^{5} - 3 x - 2 \, dx \) using a formula from geometry.
Problem 4.1182easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{-3}^{0} \left|{x + 2}\right| \, dx \) using a formula from geometry.
Problem 4.1183easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{4} \sqrt{16 - x^{2}} \, dx \) using a formula from geometry.
Problem 4.1184easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{1}^{5} - 2 x \, dx \) using a formula from geometry.
Problem 4.1185easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{0}^{2} x + 4 \, dx \) using a formula from geometry.
Problem 4.1186easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{4} 2 x + 2 \, dx \) using a formula from geometry.
Problem 4.1187easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-1}^{2} 2 x + 2 \, dx \) using a formula from geometry.
Problem 4.1188easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{-3}^{3} \sqrt{9 - x^{2}} \, dx \) using a formula from geometry.
Problem 4.1189easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{5} x + 2 \, dx \) using a formula from geometry.
Problem 4.1193easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-3}^{2} \left|{x}\right| \, dx \) using a formula from geometry.
Problem 4.1194easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{0}^{1} 2 - 2 x \, dx \) using a formula from geometry.
Problem 4.1196easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{-4}^{4} \sqrt{16 - x^{2}} \, dx \) using a formula from geometry.
Problem 4.1268easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-5}^{5} \sqrt{25 - x^{2}} \, dx \) using a formula from geometry.
Problem 4.1269easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-4}^{3} \left|{x}\right| \, dx \) using a formula from geometry.
Problem 4.1270easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{-4}^{0} \left|{x + 2}\right| \, dx \) using a formula from geometry.
Problem 4.1271easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-4}^{2} \left|{x + 2}\right| \, dx \) using a formula from geometry.
Problem 4.1272easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{-1}^{3} \left|{x}\right| \, dx \) using a formula from geometry.
Problem 4.1273easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{1}^{3} 3 - 3 x \, dx \) using a formula from geometry.
Problem 4.1274easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{3}^{4} 4 - 2 x \, dx \) using a formula from geometry.
Problem 4.1275easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{2}^{3} 6 - 2 x \, dx \) using a formula from geometry.
Problem 4.1276easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{0}^{5} \left|{x - 1}\right| \, dx \) using a formula from geometry.
Problem 4.1277easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{2}^{5} - x \, dx \) using a formula from geometry.
Problem 4.1367easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{4}^{5} 3 - x \, dx \) using a formula from geometry.
Problem 4.1368easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{1}^{4} - x \, dx \) using a formula from geometry.
Problem 4.1370easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{0}^{4} 6 - x \, dx \) using a formula from geometry.
Problem 4.1372easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{3}^{5} 2 x + 1 \, dx \) using a formula from geometry.
Problem 4.1373easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{-3}^{2} \left|{x + 1}\right| \, dx \) using a formula from geometry.
Problem 4.1374easy Video✓ Nihil obstat
Evaluate \( \displaystyle \int_{2}^{3} - 3 x - 1 \, dx \) using a formula from geometry.
Problem 4.1375easy Video✓ Every equation proved
Evaluate \( \displaystyle \int_{0}^{2} - x \, dx \) using a formula from geometry.
Problem 4.1376easy Video✓ Every equation proved