Applying Properties of Definite Integrals
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AP Calculus AB › Applying Properties of Definite Integrals
Given $\int_{2}^{6} t(x),dx=3$ and $\int_{6}^{10} t(x),dx=3$, find $\int_{10}^{2} t(x),dx$.
$0$
$-3$
$6$
$-6$
$3$
Explanation
This question assesses the skill of applying properties of definite integrals. Reversal makes the integral from 10 to 2 the negative of from 2 to 10. From 2 to 10 is 3 + 3 = 6, so -6. Additivity builds the larger interval first. A tempting distractor is 6, which forgets the reversal 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.
If $\int_{-1}^{1} h(x),dx=4$, what is $\int_{-1}^{1} \left(3h(x)+2\right),dx$?
$-8$
$12$
$16$
$8$
$-16$
Explanation
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 34 + 22 = 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.
Given $\int_{1}^{2} h(x),dx=4$ and $\int_{2}^{5} h(x),dx=4$, find $\int_{1}^{5} h(x),dx$.
$8$
$4$
$16$
$-8$
$0$
Explanation
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.
If $\int_{2}^{7} r(x),dx=9$ and $\int_{2}^{4} r(x),dx=1$, what is $\int_{4}^{7} r(x),dx$?
$-10$
$10$
$-8$
$1$
$8$
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.
If $\int_{-2}^{1} u(x),dx=-6$, what is $\int_{-2}^{1} \left(\tfrac{2}{3}u(x)\right),dx$?
$-4$
$-9$
$4$
$9$
$-6$
Explanation
This question assesses the skill of applying properties of definite integrals. The scalar 2/3 factors out, giving (2/3) * -6 = -4. This property applies directly to the given integral. No other adjustments are needed. A tempting distractor is -9, which uses 3/2 instead of 2/3. 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.
Given $\int_{0}^{8} w(x),dx=-12$, find $\int_{0}^{8} \left(\tfrac{1}{4}w(x)\right),dx$.
$3$
$-48$
$12$
$-3$
$48$
Explanation
This question assesses the skill of applying properties of definite integrals. The scalar multiple 1/4 factors out, giving (1/4) * -12 = -3. This property simplifies the computation directly. No need for further evaluation. A tempting distractor is -48, which multiplies by 4 instead of dividing. 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 $\int_{0}^{3} v(x),dx=2$ and $\int_{0}^{8} v(x),dx=9$, what is $\int_{3}^{8} v(x),dx$?
$-11$
$7$
$9$
$11$
$-7$
Explanation
This question assesses the skill of applying properties of definite integrals. Using additivity in reverse, the integral from 3 to 8 is from 0 to 8 minus 0 to 3. That's 9 - 2 = 7. This subtracts the unwanted portion. A tempting distractor is 11, which adds 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.
Given $\int_{-2}^{3} g(x),dx=9$, what is $\int_{-2}^{3} \left(g(x)-g(x)\right),dx$?
$18$
$-18$
$-9$
$9$
$0$
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.
If $\int_{3}^{7} q(x),dx=2$, what is $\int_{7}^{3} \left(2q(x)-1\right),dx$?
$-4$
$4$
$8$
$0$
$-8$
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.
Given $\int_{0}^{2} p(x),dx=-4$ and $\int_{2}^{4} p(x),dx=9$, find $\int_{0}^{4} p(x),dx$.
$13$
$-5$
$-13$
$9$
$5$
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.