What is the integral from 0 to 2 of 9x^2e^{x^3} ?

The integral from 0 to 2 of 9x^2e^{x^3} is 3(e^8-1)

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 9x^2e^{x^3}

\int_{0}^{2} 9 x^{2} e^{x^{3}} d x

3 (e^{8} - 1)

Decimal

8939.87396 \dots

Solution steps

\int_{0}^{2} 9 x^{2} e^{x^{3}} d x

Take the constant out:

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

= 9 \cdot \int_{0}^{2} x^{2} e^{x^{3}} d x

Apply u-substitution

: \int_{0}^{8} \frac{e ^{u}}{3} d u

= 9 \cdot \int_{0}^{8} \frac{e ^{u}}{3} d u

Take the constant out:

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

= 9 \cdot \frac{1}{3} \cdot \int_{0}^{8} e^{u} d u

Use the common integral:

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

= 9 \cdot \frac{1}{3} [e^{u}]_{0}^{8}

Simplify

9 \cdot \frac{1}{3} [e^{u}]_{0}^{8} : 3 [e^{u}]_{0}^{8}

= 3 [e^{u}]_{0}^{8}

Compute the boundaries:

e^{8} - 1

= 3 (e^{8} - 1)

\int_{1}^{4} (\frac{1}{x} + x) d x

\int_{- 1}^{1} \frac{y ^{3}}{3} + 2 y^{3} d y

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

\int_{6}^{5} (sin (x)) d x

\int_{2}^{8} - \frac{6}{x} d x

What is the integral from 0 to 2 of 9x^2e^{x^3} ?
The integral from 0 to 2 of 9x^2e^{x^3} is 3(e^8-1)