# How do you integrate \int _ { 0} ^ { \pi } ( 8e ^ { x } + 6\sin ( x ) ) d x?

Sep 3, 2017

${\int}_{0}^{\pi} \left(8 {e}^{x} + 6 \sin \left(x\right)\right) . \mathrm{dx} = 177$

#### Explanation:

If we look at integrating the two sections: $8 {e}^{x}$ and $6 \sin \left(x\right)$.

Lets look at ${\int}_{0}^{\pi} 8 {e}^{x} . \mathrm{dx}$. We can integrate the exponential function by dividing the coefficient of e by the derivative of the exponent.

The derivative of the exponent is:
$f \left(x\right) = x$
$f ' \left(x\right) = 1$

So divide by 1. Which means (before considering limits):
$\int 8 {e}^{x} \mathrm{dx}$
$= 8 {e}^{x} + c$

Then lets look at the other section of the integral.
$6 \sin \left(x\right)$

Note that $\int \sin \left(x\right) \mathrm{dx} = - \cos x + C$. This is because the derivative of $- \cos \left(x\right)$ is $\sin \left(x\right)$.

So:
$\int 6 \sin \left(x\right) \mathrm{dx}$
$= - 6 \cos \left(x\right) + c$

Now add these two sections together and apply the limits of $0$ and $\pi$.

${\left[8 {e}^{x} - 6 \sin \left(x\right) + c\right]}^{\pi}$

Then substitute the upper limit as x then subtract the equation with the lower limit substituted.

$\left[8 {e}^{\pi} - 6 \sin \left(\pi\right) + \cancel{c}\right] - \left[8 {e}^{0} - 6 \sin \left(0\right) + \cancel{c}\right]$

Note that $\sin \left(0\right) = \sin \left(\pi\right) = 0$ but ${e}^{0} = 1$:

$= 8 \left({e}^{\pi} - 1\right)$

$= 177.1$