∫0−1(e3x+7x−5)dx
=∫0−1e3xdx+7∫0−1xdx−5∫0−1dx
[e3x3]0−1+[7x22]0−1−5[x]0−1
=e−3∗03−e−33+7x∗02−7(−1)22−5(0−(−1))
=13−13e3−72−5
=2−21−306−13e3
=−496−13e3
=−98e3−26e3
Evaluate: ∫π20sin2xsinx+cosxdx.
OR
Evaluate: ∫2−1(e3x+7x−5)dx as a limit of sums.