What is the integral from 0 to 1 of 5x(x^2-1)^8 ?

The integral from 0 to 1 of 5x(x^2-1)^8 is 5/18

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 5x(x^2-1)^8

\int_{0}^{1} 5 x (x^{2} - 1)^{8} d x

\frac{5}{18}

Decimal

0.27777 \dots

Solution steps

\int_{0}^{1} 5 x (x^{2} - 1)^{8} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} x (x^{2} - 1)^{8} d x

Apply u-substitution

: \int_{- 1}^{0} \frac{u ^{8}}{2} d u

= 5 \cdot \int_{- 1}^{0} \frac{u ^{8}}{2} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int_{- 1}^{0} u^{8} d u

Apply the Power Rule

: [\frac{u ^{9}}{9}]_{- 1}^{0}

= 5 \cdot \frac{1}{2} [\frac{u ^{9}}{9}]_{- 1}^{0}

Simplify

5 \cdot \frac{1}{2} [\frac{u ^{9}}{9}]_{- 1}^{0} : \frac{5}{2} [\frac{u ^{9}}{9}]_{- 1}^{0}

= \frac{5}{2} [\frac{u ^{9}}{9}]_{- 1}^{0}

Compute the boundaries:

\frac{1}{9}

= \frac{5}{2} \cdot \frac{1}{9}

Simplify

= \frac{5}{18}

\int_{0}^{6} (ln (x))^{2} d x

\int_{0}^{12} \frac{x}{x + 4} d x

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

\int_{- 1}^{18} e^{- \frac{x}{2}} d x

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

What is the integral from 0 to 1 of 5x(x^2-1)^8 ?
The integral from 0 to 1 of 5x(x^2-1)^8 is 5/18