What is the integral from 0 to 2 of (x-1)e^{(x-1)^2} ?

The integral from 0 to 2 of (x-1)e^{(x-1)^2} is 0

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integral from 0 to 2 of (x-1)e^{(x-1)^2}

\int_{0}^{2} (x - 1) e^{(x - 1)^{2}} d x

0

Solution steps

\int_{0}^{2} (x - 1) e^{(x - 1)^{2}} d x

Apply u-substitution

: \int_{- 1}^{1} e^{u^{2}} u d u

= \int_{- 1}^{1} e^{u^{2}} u d u

Apply Integration By Parts

: [\frac{1}{2} e^{u^{2}} u^{2} - \int e^{u^{2}} u^{3} d u]_{- 1}^{1}

= [\frac{1}{2} e^{u^{2}} u^{2} - \int e^{u^{2}} u^{3} d u]_{- 1}^{1}

\int e^{u^{2}} u^{3} d u = \frac{1}{2} (e^{u^{2}} u^{2} - e^{u^{2}})

= [\frac{1}{2} e^{u^{2}} u^{2} - \frac{1}{2} (e^{u^{2}} u^{2} - e^{u^{2}})]_{- 1}^{1}

Simplify

[\frac{1}{2} e^{u^{2}} u^{2} - \frac{1}{2} (e^{u^{2}} u^{2} - e^{u^{2}})]_{- 1}^{1} : [\frac{1}{2} e^{u^{2}}]_{- 1}^{1}

= [\frac{1}{2} e^{u^{2}}]_{- 1}^{1}

Compute the boundaries:

0

= 0

f (x) = (2 x - 5)^{3}, at x = 1

\int e^{8 x} cos (3 x) d x

f (x) = 6 tan (x), at x = \frac{π}{3}

\frac{\partial}{\partial x} (2 x - 1)

\int x y + 1 d y

What is the integral from 0 to 2 of (x-1)e^{(x-1)^2} ?
The integral from 0 to 2 of (x-1)e^{(x-1)^2} is 0