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

1 Answer
Apr 16, 2016

(e^5-1)/5

Explanation:

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