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How do you find the definite integral for: # e^(5x) dx# for the intervals #[0, 1]#?

Answer 1

Charles Anctil

#(e^5-1)/5#

We want to find:

#int_0^1e^(5x)dx#

Our goal for integration should be to get this integral into the pattern:

#inte^udu=e^u+C#

Thus, we substitute and let #u=5x#, so that #(du)/dx=5# and #du=5dx#.

To have our #du=5dx# value inside the initial integral, we will have to multiply the interior of the integral by #5#. Balance this by multiplying the exterior by #1/5#.

#=1/5int_0^1e^(5x)*5dx#

We now see that this will fit the #inte^udu# mold. However, be careful when substituting these in--since the integrand has changed from #dx# to #du#, we will have to change the bounds of integration as well.

Do this by plugging the current bounds of #0# and #1# into the equation for #u#, #u=5x#.

#u(0)=5(0)=0# #u(1)=5(1)=5#

Thus,

#1/5int_0^1e^(5x)*5dx=1/5int_0^5e^udu#

We can now evaluate the integral from #0# to #5#:

#=1/5(e^u)]_0^5=1/5(e^5-e^0)=(e^5-1)/5#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.