Applying Properties of Definite Integrals - AP Calculus AB

Given $\int_{0}^{2} p(x)\,dx=-4$ and $\int_{2}^{4} p(x)\,dx=9$ , find $\int_{0}^{4} p(x)\,dx$ .
Answer: $5$ 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.
If $\int_{2}^{9} m(x)\,dx=-4$ , what is $\int_{2}^{9} \left(m(x)+5m(x)\right)\,dx$ ?
Answer: $-24$ This problem requires applying the linearity property of definite integrals to combine like terms. Given $\int_{2}^{9} m(x)\,dx = -4$ , we need $\int_{2}^{9} (m(x) + 5m(x))\,dx$ . First, simplify the integrand: $m(x) + 5m(x) = 6m(x)$ . Then use the constant multiple rule: $\int_{2}^{9} 6m(x)\,dx = 6\int_{2}^{9} m(x)\,dx = 6(-4) = -24$ . A tempting mistake is to integrate each term separately as $\int m(x)\,dx + 5\int m(x)\,dx = -4 + 5(-4) = -24$ , which gives the correct answer but takes extra steps. Always simplify the integrand first, then apply integral properties for efficiency.
Given $\int_{0}^{5} h(x)\,dx=2$ and $\int_{5}^{9} h(x)\,dx=11$ , find $\int_{0}^{9} h(x)\,dx$ .
Answer: $13$ This problem requires applying the additivity property of definite integrals to combine adjacent intervals. Since the intervals $[0,5]$ and $[5,9]$ share the common endpoint 5, we can apply the additivity property: $\int_{0}^{9} h(x)\,dx = \int_{0}^{5} h(x)\,dx + \int_{5}^{9} h(x)\,dx$ . Substituting the given values: $\int_{0}^{9} h(x)\,dx = 2 + 11 = 13$ . 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.
If $\int_{-1}^{4} g(x)\,dx=9$ , what is $\int_{4}^{-1} 3g(x)\,dx$ ?
Answer: $-27$ This problem requires applying properties of definite integrals to find the value without direct evaluation. We need to find $\int_{4}^{-1} 3g(x)\,dx$ given that $\int_{-1}^{4} g(x)\,dx = 9$ . First, we apply the constant multiple property: $\int_{4}^{-1} 3g(x)\,dx = 3\int_{4}^{-1} g(x)\,dx$ . Next, we use the reversal property: $\int_{4}^{-1} g(x)\,dx = -\int_{-1}^{4} g(x)\,dx = -9$ . Therefore, $\int_{4}^{-1} 3g(x)\,dx = 3(-9) = -27$ . 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.
Given $\int_{1}^{2} h(x)\,dx=4$ and $\int_{2}^{5} h(x)\,dx=4$ , find $\int_{1}^{5} h(x)\,dx$ .
Answer: $8$ This question assesses the skill of applying properties of definite integrals. The additivity property allows us to combine the integrals from 1 to 2 and 2 to 5 into the integral from 1 to 5. Both given integrals are 4, so their sum is 8. This direct addition determines the value efficiently. A tempting distractor is 16, which might come from mistakenly multiplying the values 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.
Given $\int_{0}^{4} r(x)\,dx=1$ and $\int_{4}^{10} r(x)\,dx=7$ , what is $\int_{10}^{0} r(x)\,dx$ ?
Answer: $-8$ This problem combines the additive property and reversal property of definite integrals. We know $\int_{0}^{4} r(x)\,dx = 1$ and $\int_{4}^{10} r(x)\,dx = 7$ , and need $\int_{10}^{0} r(x)\,dx$ . First, find $\int_{0}^{10} r(x)\,dx$ using additivity: $\int_{0}^{10} r(x)\,dx = \int_{0}^{4} r(x)\,dx + \int_{4}^{10} r(x)\,dx = 1 + 7 = 8$ . Then apply the reversal property: $\int_{10}^{0} r(x)\,dx = -\int_{0}^{10} r(x)\,dx = -8$ . A common error is reversing only one of the given integrals, yielding incorrect results like 6. Remember the checklist: additivity combines adjacent intervals, reversal changes sign, and work systematically.
If $\int_{0}^{10} p(x)\,dx=13$ , what is $\int_{0}^{10} \left(p(x)-1\right)\,dx$ ?
Answer: $3$ 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.
If $\int_{2}^{7} r(x)\,dx=9$ and $\int_{2}^{4} r(x)\,dx=1$ , what is $\int_{4}^{7} r(x)\,dx$ ?
Answer: $8$ 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.
If $\int_{-1}^{1} h(x)\,dx=4$ , what is $\int_{-1}^{1} \left(3h(x)+2\right)\,dx$ ?
Answer: $16$ This question assesses the skill of applying properties of definite integrals. Linearity splits the integral into 3 times the integral of h plus the integral of 2. That's 3*4 + 2*2 = 12 + 4 = 16. Constant integration uses the interval length of 2. A tempting distractor is 8, 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.
If $\int_{1}^{4} v(x)\,dx=3$ and $\int_{4}^{9} v(x)\,dx=12$ , find $\int_{9}^{1} v(x)\,dx$ .
Answer: $-15$ This problem tests your ability to apply both the additivity and reversal properties of definite integrals. First, find $\int_{1}^{9} v(x)\,dx$ using additivity: $\int_{1}^{9} v(x)\,dx = \int_{1}^{4} v(x)\,dx + \int_{4}^{9} v(x)\,dx = 3 + 12 = 15$ . Then apply the reversal property: $\int_{9}^{1} v(x)\,dx = -\int_{1}^{9} v(x)\,dx = -15$ . 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.
If $\int_{0}^{3} s(x)\,dx=2$ and $\int_{3}^{9} s(x)\,dx=5$ , find $\int_{0}^{9} [s(x)-1] \,dx$ .
Answer: $-2$ This problem requires applying properties of definite integrals to find the value without direct evaluation. We need to find $\int_{0}^{9} [s(x)-1]\,dx$ given $\int_{0}^{3} s(x)\,dx = 2$ and $\int_{3}^{9} s(x)\,dx = 5$ . First, we split the integral: $\int_{0}^{9} [s(x)-1]\,dx = \int_{0}^{9} s(x)\,dx - \int_{0}^{9} 1\,dx$ . Using the additive property: $\int_{0}^{9} s(x)\,dx = \int_{0}^{3} s(x)\,dx + \int_{3}^{9} s(x)\,dx = 2 + 5 = 7$ . The integral of 1 over [0,9] is 9, so $\int_{0}^{9} [s(x)-1]\,dx = 7 - 9 = -2$ . A common error is forgetting to integrate the constant -1 over the entire interval, which would give 7 (choice A). Remember the properties checklist: reversal changes sign, splitting uses addition, and constants factor out.
If $\int_{0}^{5} p(x)\,dx=-7$ , what is $\int_{0}^{5} [p(x)+2] \,dx$ ?
Answer: $3$ This problem requires applying properties of definite integrals to find the value without direct evaluation. We need to find $\int_{0}^{5} [p(x)+2]\,dx$ given that $\int_{0}^{5} p(x)\,dx = -7$ . Using the linearity property, we can split the integral: $\int_{0}^{5} [p(x)+2]\,dx = \int_{0}^{5} p(x)\,dx + \int_{0}^{5} 2\,dx$ . The first integral equals -7, and the second integral equals $2(5-0) = 10$ . Therefore, $\int_{0}^{5} [p(x)+2]\,dx = -7 + 10 = 3$ . 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.
Given $\int_{-3}^{2} r(x)\,dx=-6$ , what is $\int_{-3}^{2} [-r(x)]\,dx$ ?
Answer: $6$ This problem requires applying properties of definite integrals to find the value without direct evaluation. We need to find $\int_{-3}^{2} [-r(x)]\,dx$ given that $\int_{-3}^{2} r(x)\,dx = -6$ . Using the constant multiple property with $c = -1$ : $\int_{-3}^{2} [-r(x)]\,dx = -\int_{-3}^{2} r(x)\,dx$ . Since $\int_{-3}^{2} r(x)\,dx = -6$ , we have $\int_{-3}^{2} [-r(x)]\,dx = -(-6) = 6$ . 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.
If $\int_{-3}^{1} k(x)\,dx=2$ and $\int_{1}^{6} k(x)\,dx=-9$ , find $\int_{-3}^{6} k(x)\,dx$ .
Answer: $-7$ This problem tests the additive property of definite integrals over adjacent intervals. Given $\int_{-3}^{1} k(x)\,dx = 2$ and $\int_{1}^{6} k(x)\,dx = -9$ , we need $\int_{-3}^{6} k(x)\,dx$ . Since the intervals [-3,1] and [1,6] are adjacent at x=1, we can use additivity: $\int_{-3}^{6} k(x)\,dx = \int_{-3}^{1} k(x)\,dx + \int_{1}^{6} k(x)\,dx = 2 + (-9) = -7$ . A tempting error is to subtract the integrals thinking of it as "from -3 to 1, then from 1 to 6," which would give 2 - (-9) = 11. Always remember: additivity means adding the integrals, regardless of their signs.
Given $\int_{-1}^{4} q(x)\,dx=-2$ , find $\int_{-1}^{4} \left(-2q(x)\right)\,dx$ .
Answer: $4$ 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.
Given $\int_{-1}^{2} s(x)\,dx=-6$ , find $\int_{-1}^{2} \big(2s(x)-3\big)\,dx$ .
Answer: $-21$ This problem requires applying the linearity property of definite integrals to handle both a constant multiple and a constant term. Using linearity: $\int_{-1}^{2} \big(2s(x)-3\big)\,dx = \int_{-1}^{2} 2s(x)\,dx - \int_{-1}^{2} 3\,dx = 2\int_{-1}^{2} s(x)\,dx - 3(2-(-1))$ . Substituting the given value: $2(-6) - 3(3) = -12 - 9 = -21$ . The constant 3 integrates to 3 times the interval length, which is 3. A common mistake would be to write -3 instead of -9 for the constant term, forgetting to multiply by the interval length. Remember the properties checklist: linearity handles each term separately, constants factor out, and constant functions integrate to constant times interval length.
If $\int_{2}^{8} p(x)\,dx=-3$ , what is $\int_{2}^{8} \big(p(x)+5\big)\,dx$ ?
Answer: $27$ This problem tests your understanding of the linearity property of definite integrals, specifically how constants affect integration. We can split the integral using linearity: $\int_{2}^{8} \big(p(x)+5\big)\,dx = \int_{2}^{8} p(x)\,dx + \int_{2}^{8} 5\,dx$ . The first integral equals -3 (given), and the second integral of the constant 5 over the interval $[2,8]$ equals $5(8-2) = 30$ . Therefore, the answer is $-3 + 30 = 27$ . A common error would be to simply add 5 to -3, giving 2, without recognizing that we're integrating the constant 5 over the entire interval. Always remember the properties checklist: linearity allows splitting sums, constants integrate to constant times interval length, and each term must be integrated separately.
Given $\int_{-5}^{0} u(x)\,dx=9$ and $\int_{0}^{2} u(x)\,dx=-4$ , find $\int_{-5}^{2} u(x)\,dx$ .
Answer: $5$ This problem requires applying the additivity property of definite integrals to combine two adjacent intervals. Since the intervals $[-5,0]$ and $[0,2]$ share the common endpoint 0, we can use additivity: $\int_{-5}^{2} u(x)\,dx = \int_{-5}^{0} u(x)\,dx + \int_{0}^{2} u(x)\,dx$ . Substituting the given values: $\int_{-5}^{2} u(x)\,dx = 9 + (-4) = 5$ . 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 $-13$ or $13$ . Remember the properties checklist: additivity combines adjacent intervals by addition, signs are preserved as given, and endpoints must match for direct addition.
If $\int_{-2}^{4} g(x)\,dx=9$ , what is $\int_{4}^{-2} 3g(x)\,dx$ ?
Answer: $-27$ This problem tests applying properties of definite integrals, particularly the constant multiple rule and reversal property. We start with $\int_{-2}^{4} g(x)\,dx = 9$ and need $\int_{4}^{-2} 3g(x)\,dx$ . By the reversal property, $\int_{4}^{-2} 3g(x)\,dx = -\int_{-2}^{4} 3g(x)\,dx$ . Using the constant multiple rule, $\int_{-2}^{4} 3g(x)\,dx = 3\int_{-2}^{4} g(x)\,dx = 3(9) = 27$ . Therefore, $\int_{4}^{-2} 3g(x)\,dx = -27$ . A tempting error is to multiply by 3 but forget the negative from reversing limits, giving 27. Always check: constants factor out, reversal changes sign, and order matters.
If $\int_{5}^{9} t(x)\,dx=0$ , what is $\int_{5}^{9} \left(7t(x)\right)\,dx$ ?
Answer: $0$ 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.

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