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

The integral from 0 to 2 of 4xe^x is 4(e^2+1)

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 4xe^x

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

4 (e^{2} + 1)

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

: [e^{x} x - \int e^{x} d x]_{0}^{2}

= 4 [e^{x} x - \int e^{x} d x]_{0}^{2}

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

= 4 [e^{x} x - e^{x}]_{0}^{2}

Compute the boundaries:

e^{2} + 1

= 4 (e^{2} + 1)

x \to 1 lim (\frac{4 x + 3}{2 x + 1})

n = 1 \sum \infty 4 \cdot (\frac{4}{7})^{n}

\int (\frac{x ( x - 8 )}{x + 1})^{2} d x

f (x) = 5 x (sin (x) + cos (x))

\int sin (π + 2 y) - sin (2 y) d y

What is the integral from 0 to 2 of 4xe^x ?
The integral from 0 to 2 of 4xe^x is 4(e^2+1)