What is the integral from 0 to 1 of 7x(x^2-1)^5 ?

The integral from 0 to 1 of 7x(x^2-1)^5 is -7/12

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 7x(x^2-1)^5

\int_{0}^{1} 7 x (x^{2} - 1)^{5} d x

- \frac{7}{12}

Decimal

- 0.58333 \dots

Solution steps

\int_{0}^{1} 7 x (x^{2} - 1)^{5} d x

Take the constant out:

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

= 7 \cdot \int_{0}^{1} x (x^{2} - 1)^{5} d x

Apply u-substitution

: \int_{- 1}^{0} \frac{u ^{5}}{2} d u

= 7 \cdot \int_{- 1}^{0} \frac{u ^{5}}{2} d u

Take the constant out:

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

= 7 \cdot \frac{1}{2} \cdot \int_{- 1}^{0} u^{5} d u

Apply the Power Rule

: [\frac{u ^{6}}{6}]_{- 1}^{0}

= 7 \cdot \frac{1}{2} [\frac{u ^{6}}{6}]_{- 1}^{0}

Simplify

7 \cdot \frac{1}{2} [\frac{u ^{6}}{6}]_{- 1}^{0} : \frac{7}{2} [\frac{u ^{6}}{6}]_{- 1}^{0}

= \frac{7}{2} [\frac{u ^{6}}{6}]_{- 1}^{0}

Compute the boundaries:

- \frac{1}{6}

= \frac{7}{2} (- \frac{1}{6})

Simplify

= - \frac{7}{12}

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

\int_{1}^{2} 3 x^{2} + 4 x d x

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

\int_{0}^{2} \frac{1}{2} (e^{x} + e^{- x}) d x

\int_{0}^{π} 2 x sin (2 x^{2}) d x

What is the integral from 0 to 1 of 7x(x^2-1)^5 ?
The integral from 0 to 1 of 7x(x^2-1)^5 is -7/12