What is the integral from-1 to 1 of 4x^2(-x^3-1)^3 ?

The integral from-1 to 1 of 4x^2(-x^3-1)^3 is -16/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 1 of 4x^2(-x^3-1)^3

\int_{- 1}^{1} 4 x^{2} (- x^{3} - 1)^{3} d x

- \frac{16}{3}

Decimal

- 5.33333 \dots

Solution steps

\int_{- 1}^{1} 4 x^{2} (- x^{3} - 1)^{3} d x

Take the constant out:

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

= 4 \cdot \int_{- 1}^{1} x^{2} (- x^{3} - 1)^{3} d x

Expand

x^{2} (- x^{3} - 1)^{3} : - x^{11} - 3 x^{8} - 3 x^{5} - x^{2}

= 4 \cdot \int_{- 1}^{1} - x^{11} - 3 x^{8} - 3 x^{5} - x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (- \int_{- 1}^{1} x^{11} d x - \int_{- 1}^{1} 3 x^{8} d x - \int_{- 1}^{1} 3 x^{5} d x - \int_{- 1}^{1} x^{2} d x)

\int_{- 1}^{1} x^{11} d x = 0

\int_{- 1}^{1} 3 x^{8} d x = \frac{2}{3}

\int_{- 1}^{1} 3 x^{5} d x = 0

\int_{- 1}^{1} x^{2} d x = \frac{2}{3}

= 4 (- 0 - \frac{2}{3} - 0 - \frac{2}{3})

Simplify

4 (- 0 - \frac{2}{3} - 0 - \frac{2}{3}) : - \frac{16}{3}

= - \frac{16}{3}

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

\int_{c o s (y)}^{0} e^{x} sin (y) d x

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

\int_{0}^{y} e^{- (x + y)} d x

\int_{0}^{2} x^{2} + 4 x d x

What is the integral from-1 to 1 of 4x^2(-x^3-1)^3 ?
The integral from-1 to 1 of 4x^2(-x^3-1)^3 is -16/3