Suppose $\int_{0}^{2} f(t)=a$ , where a is a constant.Calculate $\int_{0}^{2} 5 f(2-t) d t$ .
A) $-10 a$
B) $10 a$
C) $-5 a$
D) $5 a$

Question

Quizplus · Accepted Answer

The integral $\int_{0}^{2} 5 f(2-t) d t$ can be solved by a substitution method. Let $u = 2-t$, then $du = -dt$. When $t = 0$, $u = 2$, and when $t = 2$, $u = 0$. Thus, the integral becomes $\int_{2}^{0} 5 f(u) (-du)$, which simplifies to $\int_{0}^{2} 5 f(u) du$. Since $\int_{0}^{2} f(u) du = a$, the integral is $5a$.

Suppose ∫02f(t)=a\int_{0}^{2} f(t)=a∫02​f(t)=a , Where a Is a Constant ∫025f(2−t)dt\int_{0}^{2} 5 f(2-t) d t∫02​5f(2−t)dt

Suppose $\int_{0}^{2} f(t)=a$ , Where a Is a Constant $\int_{0}^{2} 5 f(2-t) d t$