What is int_(0)^(1) e^(5x)dx 10e5xdx?

1 Answer
Oct 28, 2015

1/5(e^5-1)15(e51)

Explanation:

Using the substitution technique :

Let u=5xu=5x. Then du = 5dx =>1/5du=dxdu=5dx15du=dx

Limits : x=0=>u=0 and x=1=>u=5x=0u=0andx=1u=5

Therefore the original integral becomes

1/5int_0^5e^udu=1/5[e^u]_0^5=1/5(e^5-1)1550eudu=15[eu]50=15(e51)