What is the integral from-2 to-1 of x^2e^{x^3+8} ?

The integral from-2 to-1 of x^2e^{x^3+8} is (e^7-1)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from-2 to-1 of x^2e^{x^3+8}

\int_{- 2}^{- 1} x^{2} e^{x^{3} + 8} d x

\frac{e ^{7} - 1}{3}

Decimal

365.21105 \dots

Solution steps

\int_{- 2}^{- 1} x^{2} e^{x^{3} + 8} d x

Apply u-substitution

= \int_{0}^{7} \frac{e ^{u}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{7} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{0}^{7}

Compute the boundaries:

e^{7} - 1

= \frac{1}{3} (e^{7} - 1)

Simplify

= \frac{e ^{7} - 1}{3}

What is the integral from-2 to-1 of x^2e^{x^3+8} ?
The integral from-2 to-1 of x^2e^{x^3+8} is (e^7-1)/3