Answer
a)$\int ^{4}_{0}2e^{2x}dx=e^{2x}+c$
b)$e^{8}-1\approx 2980$
Work Step by Step
$u=2x\Rightarrow du=2dx$
a)$\int ^{4}_{0}2e^{2x}dx=\int ^{4}_{0}e^{2x}2dx=\int ^{4}_{0}e^{u}du=e^{u}+c=e^{2x}+c$
b)$\int ^{4}_{0}2e^{2x}dx=\int ^{4}_{0}e^{2x}2dx=\int ^{4}_{0}e^{u}du=e^{2x}]^{4}_{0}=e^{2\times 4}-e^{0}=e^{8}-1\approx 2980$
