Question

Find the value of the integral that converges.
∫^[infinity]_1 6e^-x dx.

Answer 1

`\int_(1)^(\infty) 6e^(-x) dx= -6 [0 -(1)/(e)] =(6)/(e)`

Because
`\lim_(x\to\infty) (1)/(e^x) = 0`

So then this integral converges and the value obtained is
`(6)/(e)`

Explanation:

For this case we want to find the following integral:

`\int_(1)^(\infty) 6e^(-x) dx`

We can take the 6 out of the integral like this:

`6 \int_(1)^(\infty) e^(-x)`

And if we solve the integral we got:

`\int_(1)^(\infty) 6e^(-x) dx= -6 e^(-x) \Big|_1^(\infty)`

`\int_(1)^(\infty) 6e^(-x) dx= -6 [\lim_(x\to\infty) (1)/(e^x) - e^(-1)]`

`\int_(1)^(\infty) 6e^(-x) dx= -6 [0 -(1)/(e)]=(6)/(e)`

Because
`\lim_(x\to\infty) (1)/(e^x) = 0`

So then this integral converges and the value obtained is
`(6)/(e)`