Solve the problem.
-Suppose that $\int _ { - 4 } ^ { - 1 } \mathrm {~g} ( \mathrm { t } ) \mathrm { dt } = 4$. Find $\int _ { - 4 } ^ { - 1 } \frac { \mathrm { g } ( \mathrm { x } ) } { 4 } \mathrm { dx }$ and $\int _ { - 1 } ^ { - 4 } - \mathrm { g } ( \mathrm { t } ) \mathrm { dt }$.
A) $- 1 ; - 4$
B) $0 ; 4$
C) $1 ; 4$
D) $1 ; - 4$

Question

Quizplus · Accepted Answer

Using the property of definite integrals, we have:

$$\int _ { - 4 } ^ { - 1 } \frac { \mathrm { g } ( \mathrm { x } ) } { 4 } \mathrm { dx } = \frac{1}{4} \int _ { - 4 } ^ { - 1 } \mathrm { g } ( \mathrm { x } )\mathrm { dx } = \frac{1}{4} \cdot 4 = 1$$

For the second integral, we can use the fact that integrating from a to b is the same as integrating from b to a with a negative sign:

$$\int _ { - 1 } ^ { - 4 } - \mathrm { g } ( \mathrm { t } ) \mathrm { dt } = -\int _ { - 4 } ^ { - 1 } \mathrm { g } ( \mathrm { t } ) \mathrm { dt } = -4$$

Therefore, the answer is (C).

Solve the Problem ∫−4−1 g(t)dt=4\int _ { - 4 } ^ { - 1 } \mathrm {~g} ( \mathrm { t } ) \mathrm { dt } = 4∫−4−1​ g(t)dt=4

Solve the Problem $\int _ { - 4 } ^ { - 1 } \mathrm {~g} ( \mathrm { t } ) \mathrm { dt } = 4$