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.

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.

This problem requires applying properties of definite integrals to find the value without direct evaluation. We need to find $\int_{3}^{6} h(x),dx$ given $\int_{1}^{6} h(x),dx = 4$ and $\int_{1}^{3} h(x),dx = -2$. Using the additive property of integrals over adjacent intervals: $\int_{1}^{6} h(x),dx = \int_{1}^{3} h(x),dx + \int_{3}^{6} h(x),dx$. Substituting the known values: $4 = -2 + \int_{3}^{6} h(x),dx$, which gives us $\int_{3}^{6} h(x),dx = 6$. 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.

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.

This question assesses the skill of applying properties of definite integrals. Additivity combines from $-5$ to $0$ and $0$ to $4$ into $-5$ to $4$. Summing $6$ and $-1$ gives $5$. This property handles the split intervals. A tempting distractor is $7$, 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.

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.

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.

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.

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.

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.