What is the integral from 1 to 2 of 4xe^{2x} ?

The integral from 1 to 2 of 4xe^{2x} is 3e^4-e^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of 4xe^{2x}

\int_{1}^{2} 4 x e^{2 x} d x

3 e^{4} - e^{2}

Decimal

156.40539 \dots

Solution steps

\int_{1}^{2} 4 x e^{2 x} d x

Take the constant out:

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

= 4 \cdot \int_{1}^{2} x e^{2 x} d x

Apply u-substitution

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

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

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int_{2}^{4} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{2}^{4}

= 4 \cdot \frac{1}{4} [e^{u} u - \int e^{u} d u]_{2}^{4}

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

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

Simplify

4 \cdot \frac{1}{4} [e^{u} u - e^{u}]_{2}^{4} : [e^{u} u - e^{u}]_{2}^{4}

= [e^{u} u - e^{u}]_{2}^{4}

Compute the boundaries:

3 e^{4} - e^{2}

= 3 e^{4} - e^{2}

\int_{- 4}^{2} 2 x + 5 d x

\int_{0}^{x^{3}} ln (t^{2} + 1) d t

\int_{- 4}^{4} \frac{1}{x ^{2}} d x

\int_{1}^{3} \frac{x ^{3} + 2}{x ^{2}} d x

\int_{1}^{a} \frac{1}{x} d x

What is the integral from 1 to 2 of 4xe^{2x} ?
The integral from 1 to 2 of 4xe^{2x} is 3e^4-e^2