What is the integral from 0 to 2 of xe^{4-x^2} ?

The integral from 0 to 2 of xe^{4-x^2} is (e^4-1)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 2 of xe^{4-x^2}

\int_{0}^{2} x e^{4 - x^{2}} d x

\frac{e ^{4} - 1}{2}

Decimal

26.79907 \dots

Solution steps

\int_{0}^{2} x e^{4 - x^{2}} d x

Apply u-substitution

= \int_{4}^{0} - \frac{e ^{u}}{2} d u

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

= - \int_{0}^{4} - \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_{0}^{4} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{2} [e^{u}]_{0}^{4})

Simplify

= \frac{1}{2} [e^{u}]_{0}^{4}

Compute the boundaries:

e^{4} - 1

= \frac{1}{2} (e^{4} - 1)

Simplify

= \frac{e ^{4} - 1}{2}

What is the integral from 0 to 2 of xe^{4-x^2} ?
The integral from 0 to 2 of xe^{4-x^2} is (e^4-1)/2