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

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Answer 1 · 2017-09-03T00:43:14

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

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#