Answer
a) 6
b) 6
Work Step by Step
a) $\int^{3}_{-1}h(r)dr= \int^{1}_{-1}h(r)dr+\int^{3}_{1}h(r)dr$
$⇒ 6=0+\int^{3}_{1}h(r)dr$
$⇒ \int^{3}_{1}h(r)dr=6$
b) $-\int ^{1}_{3}h(u)du=\int^{3}_{1}h(u)du $
$= \int^{3}_{1}h(r)dr=6$
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.