# Evaluate the integral. `int_0^1 ye^(-2y^2) dy` Please explain each step as you solve!

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### 1 Answer

Given that `f(y)= ye^(-2y^2)`

We need to find the intergral of f(y).

`==gt int_0^1 f(y) dy = int_0^1 ye^(-2y^2) dy `

`Let x = -2y^2 ==gt dx = -4y dy ==gt dy= (dx)/(-4y)`

Now we will subsitute:

`==gt int_0^1 f(y) dy = int_0^1 ye^x (dx)/(-4y) `

`= -1/4 int_0^1 e^x dx `

`= -1/4 e^x = -1/4 e^(-2y^2) `

`==gt int_0^1 f(y) dy =`

` -1/4 (e^(-2) - e^0) = -1/4 (1/e^2 - 1) ~~ 0.2162`

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