## Calculus: Early Transcendentals 8th Edition

$\frac{e^2-1}{2e}$
$\int^1_0 cosh (t) dt$ The first step is to integrate cosh(t). (Remember that $\int^b_a cosh(x)dx = sinh(x)|^b_a$ and $sinh(x) = \frac{e^x - e^{-x}}{2}$): $= [sinh(t)]|^1_0$ $= (\frac{e^t - e^{-t}}{2})|^1_0$ Next step is to plug in the limits of integration and simplify until final answer is reached: $= (\frac{e^1-e^{-1}}{2}) - (\frac{e^0 - e^0}{2})$ $= \frac{e - \frac{1}{e}}{2} - 0$ $= \frac{\frac{e^2 - 1}{e}}{2}$ $= \frac{e^2 - 1}{2e}$