Question

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

Answer 1

`int_0^(pi)(8e^x + 6sin(x)).dx = 177`

Explanation:

If we look at integrating the two sections: `8e^x` and `6sin(x)`.

Lets look at `int_0^(pi)8e^x.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(x) = x`
`f'(x) = 1`

So divide by 1. Which means (before considering limits):
`int8e^xdx`
`= 8e^x + c`

Then lets look at the other section of the integral.
`6sin(x)`

Note that `intsin(x)dx=-cosx+C`. This is because the derivative of `-cos(x)` is `sin(x)`.

So:
`int6sin(x)dx`
`= -6cos(x) + c`

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

`[8e^x - 6sin(x) +c]^(pi)`

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

`[8e^(pi) - 6sin(pi) + cancel(c)] - [8e^0 - 6sin(0) +cancel(c)]`

Note that `sin(0)=sin(pi)=0` but `e^0=1`:

`=8(e^pi-1)`

`= 177.1`