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AP Calculus AB Quiz

AP Calculus AB Quiz: Applying Properties Of Definite Integrals

Practice Applying Properties Of Definite Integrals in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

Given ∫−32r(x) dx=−6\int_{-3}^{2} r(x)\,dx=-6∫−32​r(x)dx=−6, what is ∫−32[−r(x)] dx\int_{-3}^{2} [-r(x)]\,dx∫−32​[−r(x)]dx?

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What this quiz covers

This quiz focuses on Applying Properties Of Definite Integrals, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Given ∫−32r(x) dx=−6\int_{-3}^{2} r(x)\,dx=-6∫−32​r(x)dx=−6, what is ∫−32[−r(x)] dx\int_{-3}^{2} [-r(x)]\,dx∫−32​[−r(x)]dx?

  1. −6-6−6
  2. 666 (correct answer)
  3. 000
  4. −12-12−12
  5. 121212

Explanation: This problem requires applying properties of definite integrals to find the value without direct evaluation. We need to find ∫−32[−r(x)] dx\int_{-3}^{2} [-r(x)]\,dx∫−32​[−r(x)]dx given that ∫−32r(x) dx=−6\int_{-3}^{2} r(x)\,dx = -6∫−32​r(x)dx=−6. Using the constant multiple property with c=−1c = -1c=−1: ∫−32[−r(x)] dx=−∫−32r(x) dx\int_{-3}^{2} [-r(x)]\,dx = -\int_{-3}^{2} r(x)\,dx∫−32​[−r(x)]dx=−∫−32​r(x)dx. Since ∫−32r(x) dx=−6\int_{-3}^{2} r(x)\,dx = -6∫−32​r(x)dx=−6, we have ∫−32[−r(x)] dx=−(−6)=6\int_{-3}^{2} [-r(x)]\,dx = -(-6) = 6∫−32​[−r(x)]dx=−(−6)=6. A common mistake is thinking that negating the function doesn't change the integral value, leading to -6 (choice A). Remember the properties checklist: reversal changes sign, splitting uses addition, and constants factor out.

Question 2

If ∫05p(x) dx=−7\int_{0}^{5} p(x)\,dx=-7∫05​p(x)dx=−7, what is ∫05[p(x)+2] dx\int_{0}^{5} [p(x)+2] \,dx∫05​[p(x)+2]dx?

  1. −17-17−17
  2. 333 (correct answer)
  3. −5-5−5
  4. −9-9−9
  5. 777

Explanation: This problem requires applying properties of definite integrals to find the value without direct evaluation. We need to find ∫05[p(x)+2] dx\int_{0}^{5} [p(x)+2]\,dx∫05​[p(x)+2]dx given that ∫05p(x) dx=−7\int_{0}^{5} p(x)\,dx = -7∫05​p(x)dx=−7. Using the linearity property, we can split the integral: ∫05[p(x)+2] dx=∫05p(x) dx+∫052 dx\int_{0}^{5} [p(x)+2]\,dx = \int_{0}^{5} p(x)\,dx + \int_{0}^{5} 2\,dx∫05​[p(x)+2]dx=∫05​p(x)dx+∫05​2dx. The first integral equals -7, and the second integral equals 2(5−0)=102(5-0) = 102(5−0)=10. Therefore, ∫05[p(x)+2] dx=−7+10=3\int_{0}^{5} [p(x)+2]\,dx = -7 + 10 = 3∫05​[p(x)+2]dx=−7+10=3. A common error is to add 2 to the integral value instead of integrating the constant 2 over the interval, which would give -5 (choice C). Remember the properties checklist: reversal changes sign, splitting uses addition, and constants factor out.

Question 3

Given ∫16h(x) dx=4\int_{1}^{6} h(x)\,dx=4∫16​h(x)dx=4 and ∫13h(x) dx=−2\int_{1}^{3} h(x)\,dx=-2∫13​h(x)dx=−2, find ∫36h(x) dx\int_{3}^{6} h(x)\,dx∫36​h(x)dx.

  1. −6-6−6
  2. 666 (correct answer)
  3. 222
  4. −2-2−2
  5. 000

Explanation: This problem requires applying properties of definite integrals to find the value without direct evaluation. We need to find ∫36h(x) dx\int_{3}^{6} h(x)\,dx∫36​h(x)dx given ∫16h(x) dx=4\int_{1}^{6} h(x)\,dx = 4∫16​h(x)dx=4 and ∫13h(x) dx=−2\int_{1}^{3} h(x)\,dx = -2∫13​h(x)dx=−2. Using the additive property of integrals over adjacent intervals: ∫16h(x) dx=∫13h(x) dx+∫36h(x) dx\int_{1}^{6} h(x)\,dx = \int_{1}^{3} h(x)\,dx + \int_{3}^{6} h(x)\,dx∫16​h(x)dx=∫13​h(x)dx+∫36​h(x)dx. Substituting the known values: 4=−2+∫36h(x) dx4 = -2 + \int_{3}^{6} h(x)\,dx4=−2+∫36​h(x)dx, which gives us ∫36h(x) dx=6\int_{3}^{6} h(x)\,dx = 6∫36​h(x)dx=6. A common mistake is subtracting the integrals instead of using the additive property correctly, which might lead to -6 (choice A). Remember the properties checklist: reversal changes sign, splitting uses addition, and constants factor out.

Question 4

If ∫19s(x) dx=−10\int_{1}^{9} s(x)\,dx=-10∫19​s(x)dx=−10, what is ∫19(s(x)+5) dx\int_{1}^{9} \left(s(x)+5\right)\,dx∫19​(s(x)+5)dx?

  1. −55-55−55
  2. 303030 (correct answer)
  3. −30-30−30
  4. −50-50−50
  5. 505050

Explanation: This question assesses the skill of applying properties of definite integrals. Linearity splits into the integral of s plus 5 times the integral of 1. That's -10 + 5*8 = -10 + 40 = 30. The constant uses the interval length of 8. A tempting distractor is -50, which subtracts instead of adding. When working with definite integrals, remember this checklist: additivity over adjacent intervals, scalar multiples factor out, reversing limits adds a negative sign, and constants integrate to the constant times the interval length.

Question 5

Given ∫−50r(x) dx=6\int_{-5}^{0} r(x)\,dx=6∫−50​r(x)dx=6 and ∫04r(x) dx=−1\int_{0}^{4} r(x)\,dx=-1∫04​r(x)dx=−1, what is ∫−54r(x) dx\int_{-5}^{4} r(x)\,dx∫−54​r(x)dx?

  1. −7-7−7
  2. 555 (correct answer)
  3. 777
  4. −5-5−5
  5. 111

Explanation: This question assesses the skill of applying properties of definite integrals. Additivity combines from −5-5−5 to 000 and 000 to 444 into −5-5−5 to 444. Summing 666 and −1-1−1 gives 555. This property handles the split intervals. A tempting distractor is 777, which ignores the negative sign. When working with definite integrals, remember this checklist: additivity over adjacent intervals, scalar multiples factor out, reversing limits adds a negative sign, and constants integrate to the constant times the interval length.

Question 6

If ∫59t(x) dx=0\int_{5}^{9} t(x)\,dx=0∫59​t(x)dx=0, what is ∫59(7t(x)) dx\int_{5}^{9} \left(7t(x)\right)\,dx∫59​(7t(x))dx?

  1. 777
  2. −7-7−7
  3. 000 (correct answer)
  4. 999
  5. −9-9−9

Explanation: This question assesses the skill of applying properties of definite integrals. The scalar multiple property factors out the 7, giving 7 times the integral of t(x), which is 7 * 0 = 0. This holds regardless of the function since the base integral is zero. No additional computation is required. A tempting distractor is 7, which ignores that the integral is zero. When working with definite integrals, remember this checklist: additivity over adjacent intervals, scalar multiples factor out, reversing limits adds a negative sign, and constants integrate to the constant times the interval length.

Question 7

Given ∫−14q(x) dx=−2\int_{-1}^{4} q(x)\,dx=-2∫−14​q(x)dx=−2, find ∫−14(−2q(x)) dx\int_{-1}^{4} \left(-2q(x)\right)\,dx∫−14​(−2q(x))dx.

  1. −4-4−4
  2. 444 (correct answer)
  3. −1-1−1
  4. 111
  5. 000

Explanation: This question assesses the skill of applying properties of definite integrals. The scalar multiple property allows us to factor out -2 from the integral, giving -2 times the given integral of -2. This results in -2 * -2 = 4. No further evaluation is needed due to this property. A tempting distractor is -4, which might occur by forgetting to apply the scalar to the given value correctly. When working with definite integrals, remember this checklist: additivity over adjacent intervals, scalar multiples factor out, reversing limits adds a negative sign, and constants integrate to the constant times the interval length.

Question 8

If ∫010p(x) dx=13\int_{0}^{10} p(x)\,dx=13∫010​p(x)dx=13, what is ∫010(p(x)−1) dx\int_{0}^{10} \left(p(x)-1\right)\,dx∫010​(p(x)−1)dx?

  1. 232323
  2. 333 (correct answer)
  3. 121212
  4. −3-3−3
  5. −12-12−12

Explanation: This question assesses the skill of applying properties of definite integrals. The linearity property lets us split the integral of p(x) - 1 into the integral of p(x) minus the integral of 1. The integral of 1 from 0 to 10 is 10, so 13 - 10 equals 3. This uses the constant integral rule alongside linearity. A tempting distractor is 23, which adds 10 instead of subtracting. When working with definite integrals, remember this checklist: additivity over adjacent intervals, scalar multiples factor out, reversing limits adds a negative sign, and constants integrate to the constant times the interval length.

Question 9

Given ∫−23g(x) dx=9\int_{-2}^{3} g(x)\,dx=9∫−23​g(x)dx=9, what is ∫−23(g(x)−g(x)) dx\int_{-2}^{3} \left(g(x)-g(x)\right)\,dx∫−23​(g(x)−g(x))dx?

  1. 999
  2. −9-9−9
  3. 000 (correct answer)
  4. 181818
  5. −18-18−18

Explanation: This question assesses the skill of applying properties of definite integrals. The expression g(x) - g(x) simplifies to 0, and the integral of 0 is 0. Linearity confirms this result regardless of the given value. No computation of the original integral is needed. A tempting distractor is 9, which uses the given value without simplifying the integrand. When working with definite integrals, remember this checklist: additivity over adjacent intervals, scalar multiples factor out, reversing limits adds a negative sign, and constants integrate to the constant times the interval length.

Question 10

Given ∫02p(x) dx=−4\int_{0}^{2} p(x)\,dx=-4∫02​p(x)dx=−4 and ∫24p(x) dx=9\int_{2}^{4} p(x)\,dx=9∫24​p(x)dx=9, find ∫04p(x) dx\int_{0}^{4} p(x)\,dx∫04​p(x)dx.

  1. −13-13−13
  2. 555 (correct answer)
  3. 131313
  4. −5-5−5
  5. 999

Explanation: This question assesses the skill of applying properties of definite integrals. Additivity sums the integrals from 0 to 2 and 2 to 4 into 0 to 4. That's -4 + 9 = 5. This merges the given values directly. A tempting distractor is -13, which subtracts instead of adding. When working with definite integrals, remember this checklist: additivity over adjacent intervals, scalar multiples factor out, reversing limits adds a negative sign, and constants integrate to the constant times the interval length.

Question 11

If ∫37q(x) dx=2\int_{3}^{7} q(x)\,dx=2∫37​q(x)dx=2, what is ∫73(2q(x)−1) dx\int_{7}^{3} \left(2q(x)-1\right)\,dx∫73​(2q(x)−1)dx?

  1. −4-4−4
  2. 000 (correct answer)
  3. 444
  4. 888
  5. −8-8−8

Explanation: This question assesses the skill of applying properties of definite integrals. Reversal makes the integral from 7 to 3 the negative of from 3 to 7. For (2q - 1), it's - [22 - 14] = - [4 - 4] = 0. Linearity and constants are applied inside. A tempting distractor is -4, which forgets the constant term. When working with definite integrals, remember this checklist: additivity over adjacent intervals, scalar multiples factor out, reversing limits adds a negative sign, and constants integrate to the constant times the interval length.

Question 12

Given ∫05h(x) dx=2\int_{0}^{5} h(x)\,dx=2∫05​h(x)dx=2 and ∫59h(x) dx=11\int_{5}^{9} h(x)\,dx=11∫59​h(x)dx=11, find ∫09h(x) dx\int_{0}^{9} h(x)\,dx∫09​h(x)dx.

  1. 999
  2. 131313 (correct answer)
  3. −13-13−13
  4. −9-9−9
  5. 222222

Explanation: This problem requires applying the additivity property of definite integrals to combine adjacent intervals. Since the intervals [0,5][0,5][0,5] and [5,9][5,9][5,9] share the common endpoint 5, we can apply the additivity property: ∫09h(x) dx=∫05h(x) dx+∫59h(x) dx\int_{0}^{9} h(x)\,dx = \int_{0}^{5} h(x)\,dx + \int_{5}^{9} h(x)\,dx∫09​h(x)dx=∫05​h(x)dx+∫59​h(x)dx. Substituting the given values: ∫09h(x) dx=2+11=13\int_{0}^{9} h(x)\,dx = 2 + 11 = 13∫09​h(x)dx=2+11=13. The additivity property works because the upper limit of the first integral matches the lower limit of the second integral. A common mistake would be to subtract the values instead of adding them, yielding -9. Remember the properties checklist: additivity requires matching endpoints, direction matters for signs, and adjacent intervals combine by addition.

Question 13

If ∫14v(x) dx=3\int_{1}^{4} v(x)\,dx=3∫14​v(x)dx=3 and ∫49v(x) dx=12\int_{4}^{9} v(x)\,dx=12∫49​v(x)dx=12, find ∫91v(x) dx\int_{9}^{1} v(x)\,dx∫91​v(x)dx.

  1. 151515
  2. −9-9−9
  3. 999
  4. −15-15−15 (correct answer)
  5. −12-12−12

Explanation: This problem tests your ability to apply both the additivity and reversal properties of definite integrals. First, find ∫19v(x) dx\int_{1}^{9} v(x)\,dx∫19​v(x)dx using additivity: ∫19v(x) dx=∫14v(x) dx+∫49v(x) dx=3+12=15\int_{1}^{9} v(x)\,dx = \int_{1}^{4} v(x)\,dx + \int_{4}^{9} v(x)\,dx = 3 + 12 = 15∫19​v(x)dx=∫14​v(x)dx+∫49​v(x)dx=3+12=15. Then apply the reversal property: ∫91v(x) dx=−∫19v(x) dx=−15\int_{9}^{1} v(x)\,dx = -\int_{1}^{9} v(x)\,dx = -15∫91​v(x)dx=−∫19​v(x)dx=−15. The reversal of limits introduces a negative sign, changing our result from 15 to -15. A common error would be to simply add 3 and 12 without considering the reversed limits, yielding 15 instead of -15. Always remember the properties checklist: combine adjacent intervals first, then apply reversal to change sign, and the order of limits determines the final sign.

Question 14

Given ∫02f(x) dx=5\int_{0}^{2} f(x)\,dx=5∫02​f(x)dx=5 and ∫26f(x) dx=−1\int_{2}^{6} f(x)\,dx=-1∫26​f(x)dx=−1, find ∫60f(x) dx\int_{6}^{0} f(x)\,dx∫60​f(x)dx.

  1. 444
  2. −4-4−4 (correct answer)
  3. 666
  4. −6-6−6
  5. −5-5−5

Explanation: This problem requires applying properties of definite integrals, specifically the reversal and additivity properties. To find ∫60f(x) dx\int_{6}^{0} f(x)\,dx∫60​f(x)dx, we first use the reversal property: ∫60f(x) dx=−∫06f(x) dx\int_{6}^{0} f(x)\,dx = -\int_{0}^{6} f(x)\,dx∫60​f(x)dx=−∫06​f(x)dx. Next, we apply the additivity property to find ∫06f(x) dx=∫02f(x) dx+∫26f(x) dx=5+(−1)=4\int_{0}^{6} f(x)\,dx = \int_{0}^{2} f(x)\,dx + \int_{2}^{6} f(x)\,dx = 5 + (-1) = 4∫06​f(x)dx=∫02​f(x)dx+∫26​f(x)dx=5+(−1)=4. Therefore, ∫60f(x) dx=−4\int_{6}^{0} f(x)\,dx = -4∫60​f(x)dx=−4. A common error would be to simply add the given integrals without considering the direction of integration, yielding 4 instead of -4. Remember the key properties checklist: reversal changes sign, additivity requires matching endpoints, and constants factor out.

Question 15

Given ∫−50u(x) dx=9\int_{-5}^{0} u(x)\,dx=9∫−50​u(x)dx=9 and ∫02u(x) dx=−4\int_{0}^{2} u(x)\,dx=-4∫02​u(x)dx=−4, find ∫−52u(x) dx\int_{-5}^{2} u(x)\,dx∫−52​u(x)dx.

  1. 131313
  2. −13-13−13
  3. 555 (correct answer)
  4. −5-5−5
  5. 363636

Explanation: This problem requires applying the additivity property of definite integrals to combine two adjacent intervals. Since the intervals [−5,0][-5,0][−5,0] and [0,2][0,2][0,2] share the common endpoint 0, we can use additivity: ∫−52u(x) dx=∫−50u(x) dx+∫02u(x) dx\int_{-5}^{2} u(x)\,dx = \int_{-5}^{0} u(x)\,dx + \int_{0}^{2} u(x)\,dx∫−52​u(x)dx=∫−50​u(x)dx+∫02​u(x)dx. Substituting the given values: ∫−52u(x) dx=9+(−4)=5\int_{-5}^{2} u(x)\,dx = 9 + (-4) = 5∫−52​u(x)dx=9+(−4)=5. The additivity property allows us to combine these integrals directly because they form a continuous interval from -5 to 2. A common mistake would be to subtract the integrals or change signs unnecessarily, perhaps getting −13-13−13 or 131313. Remember the properties checklist: additivity combines adjacent intervals by addition, signs are preserved as given, and endpoints must match for direct addition.

Question 16

If ∫−31g(x) dx=7\int_{-3}^{1} g(x)\,dx=7∫−31​g(x)dx=7, what is ∫1−34g(x) dx\int_{1}^{-3} 4g(x)\,dx∫1−3​4g(x)dx?

  1. 282828
  2. −7-7−7
  3. −28-28−28 (correct answer)
  4. 777
  5. 212121

Explanation: This problem tests your ability to apply properties of definite integrals, specifically the constant multiple rule and reversal property. First, we apply the constant multiple property: ∫1−34g(x) dx=4∫1−3g(x) dx\int_{1}^{-3} 4g(x)\,dx = 4\int_{1}^{-3} g(x)\,dx∫1−3​4g(x)dx=4∫1−3​g(x)dx. Next, we use the reversal property: ∫1−3g(x) dx=−∫−31g(x) dx=−7\int_{1}^{-3} g(x)\,dx = -\int_{-3}^{1} g(x)\,dx = -7∫1−3​g(x)dx=−∫−31​g(x)dx=−7. Therefore, ∫1−34g(x) dx=4(−7)=−28\int_{1}^{-3} 4g(x)\,dx = 4(-7) = -28∫1−3​4g(x)dx=4(−7)=−28. A tempting error would be to multiply 4 by 7 directly without considering the reversed limits, giving 28 instead of -28. Always check your properties checklist: constants factor out, reversal changes sign, and order of limits matters.

Question 17

If ∫−14g(x) dx=9\int_{-1}^{4} g(x)\,dx=9∫−14​g(x)dx=9, what is ∫4−13g(x) dx\int_{4}^{-1} 3g(x)\,dx∫4−1​3g(x)dx?

  1. 272727
  2. −27-27−27 (correct answer)
  3. −3-3−3
  4. 333
  5. −9-9−9

Explanation: This problem requires applying properties of definite integrals to find the value without direct evaluation. We need to find ∫4−13g(x) dx\int_{4}^{-1} 3g(x)\,dx∫4−1​3g(x)dx given that ∫−14g(x) dx=9\int_{-1}^{4} g(x)\,dx = 9∫−14​g(x)dx=9. First, we apply the constant multiple property: ∫4−13g(x) dx=3∫4−1g(x) dx\int_{4}^{-1} 3g(x)\,dx = 3\int_{4}^{-1} g(x)\,dx∫4−1​3g(x)dx=3∫4−1​g(x)dx. Next, we use the reversal property: ∫4−1g(x) dx=−∫−14g(x) dx=−9\int_{4}^{-1} g(x)\,dx = -\int_{-1}^{4} g(x)\,dx = -9∫4−1​g(x)dx=−∫−14​g(x)dx=−9. Therefore, ∫4−13g(x) dx=3(−9)=−27\int_{4}^{-1} 3g(x)\,dx = 3(-9) = -27∫4−1​3g(x)dx=3(−9)=−27. A tempting error is to multiply first without considering the reversed limits, which would give 27 (choice A). Remember the properties checklist: reversal changes sign, splitting uses addition, and constants factor out.

Question 18

Given ∫13u(x) dx=2\int_{1}^{3} u(x)\,dx=2∫13​u(x)dx=2 and ∫36u(x) dx=2\int_{3}^{6} u(x)\,dx=2∫36​u(x)dx=2, find ∫165u(x) dx\int_{1}^{6} 5u(x)\,dx∫16​5u(x)dx.

  1. 444
  2. 101010
  3. 202020 (correct answer)
  4. −20-20−20
  5. −10-10−10

Explanation: This question assesses the skill of applying properties of definite integrals. Additivity combines the integrals from 1 to 3 and 3 to 6 into 1 to 6, summing to 4. The scalar 5 factors out, giving 5 * 4 = 20. These properties allow direct calculation. A tempting distractor is 10, which forgets the scalar multiple. When working with definite integrals, remember this checklist: additivity over adjacent intervals, scalar multiples factor out, reversing limits adds a negative sign, and constants integrate to the constant times the interval length.

Question 19

If ∫−6−2g(x) dx=11\int_{-6}^{-2} g(x)\,dx=11∫−6−2​g(x)dx=11, what is ∫−2−63g(x) dx\int_{-2}^{-6} 3g(x)\,dx∫−2−6​3g(x)dx?

  1. 333333
  2. −33-33−33 (correct answer)
  3. −11-11−11
  4. 111111
  5. 000

Explanation: This question assesses the skill of applying properties of definite integrals. The reversal property indicates that the integral from -2 to -6 of g(x) dx is the negative of the integral from -6 to -2. Multiplying by the scalar 3 factors out, so it's 3 times -11, yielding -33. These properties combine to find the value without direct computation. A tempting distractor is -11, which forgets the scalar multiple of 3. When working with definite integrals, remember this checklist: additivity over adjacent intervals, scalar multiples factor out, reversing limits adds a negative sign, and constants integrate to the constant times the interval length.

Question 20

If ∫27r(x) dx=9\int_{2}^{7} r(x)\,dx=9∫27​r(x)dx=9 and ∫24r(x) dx=1\int_{2}^{4} r(x)\,dx=1∫24​r(x)dx=1, what is ∫47r(x) dx\int_{4}^{7} r(x)\,dx∫47​r(x)dx?

  1. 101010
  2. 888 (correct answer)
  3. −8-8−8
  4. −10-10−10
  5. 111

Explanation: This question assesses the skill of applying properties of definite integrals. The additivity property in reverse lets us find the integral from 4 to 7 by subtracting the integral from 2 to 4 from 2 to 7. Thus, 9 - 1 equals 8. This subtraction isolates the desired interval. A tempting distractor is 10, which adds 1 instead of subtracting. When working with definite integrals, remember this checklist: additivity over adjacent intervals, scalar multiples factor out, reversing limits adds a negative sign, and constants integrate to the constant times the interval length.