What is the integral from 0 to 2 of 7xe^{4-x^2} ?

The integral from 0 to 2 of 7xe^{4-x^2} is (7(e^4-1))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of 7xe^{4-x^2}

\int_{0}^{2} 7 x e^{4 - x^{2}} d x

\frac{7 ( e ^{4} - 1 )}{2}

Decimal Notation

187.59352 \dots

Solution steps

\int_{0}^{2} 7 x e^{4 - x^{2}} d x

Take the constant out:

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

= 7 \cdot \int_{0}^{2} x e^{4 - x^{2}} d x

Apply u-substitution

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

= 7 \cdot \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

= 7 (- \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

= 7 (- (- \frac{1}{2} \cdot \int_{0}^{4} e^{u} d u))

Use the common integral:

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

= 7 (- (- \frac{1}{2} [e^{u}]_{0}^{4}))

Simplify

7 (- (- \frac{1}{2} [e^{u}]_{0}^{4})) : \frac{7}{2} [e^{u}]_{0}^{4}

= \frac{7}{2} [e^{u}]_{0}^{4}

Compute the boundaries:

e^{4} - 1

= \frac{7}{2} (e^{4} - 1)

Simplify

= \frac{7 ( e ^{4} - 1 )}{2}

\int_{2}^{5} x e^{x} d x

\int_{4}^{9} \frac{2 x - 5}{4} d x

\int_{- 1}^{0} 9 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

\int_{0}^{4} (\frac{1}{5}) e^{- \frac{x}{5}} d x

\int_{0}^{π} e^{2 x} cos (x) d x

What is the integral from 0 to 2 of 7xe^{4-x^2} ?
The integral from 0 to 2 of 7xe^{4-x^2} is (7(e^4-1))/2