What is the integral from-1 to 2 of xe^{-x^2+1} ?

The integral from-1 to 2 of xe^{-x^2+1} is (e^3-1)/(2e^3)

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integral from-1 to 2 of xe^{-x^2+1}

\int_{- 1}^{2} x e^{- x^{2} + 1} d x

\frac{e ^{3} - 1}{2 e ^{3}}

Decimal

0.47510 \dots

Solution steps

\int_{- 1}^{2} x e^{- x^{2} + 1} d x

Apply u-substitution

: \int_{0}^{- 3} - \frac{e ^{u}}{2} d u

= \int_{0}^{- 3} - \frac{e ^{u}}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 3}^{0} - \frac{e ^{u}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{2} \cdot \int_{- 3}^{0} e^{u} d u)

Use the common integral:

\int e^{u} d u = e^{u}

= - (- \frac{1}{2} [e^{u}]_{- 3}^{0})

Simplify

= \frac{1}{2} [e^{u}]_{- 3}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{3}}

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

Simplify

= \frac{e ^{3} - 1}{2 e ^{3}}

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\int_{0}^{25} 85 - 3 25 - t d t

\int_{0}^{x} \frac{t ^{2}}{t ^{4} + 1} d t

What is the integral from-1 to 2 of xe^{-x^2+1} ?
The integral from-1 to 2 of xe^{-x^2+1} is (e^3-1)/(2e^3)