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

The integral from-1 to 3 of 4x(1-x^2) is -64

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- 64

Solution steps

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

Take the constant out:

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

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

Expand

x (1 - x^{2}) : x - x^{3}

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

\int_{- 1}^{3} x d x = 4

\int_{- 1}^{3} x^{3} d x = 20

= 4 (4 - 20)

Simplify

= - 64

\int_{3}^{5} \frac{1}{x ^{2}} d x

\int_{\frac{π}{6}}^{\frac{π}{2}} x csc^{2} (x) d x

\int_{- 2}^{0} (4 x^{2} + 4 x) d x

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

\int_{0}^{1} 16 x^{3} ln (x) d x

What is the integral from-1 to 3 of 4x(1-x^2) ?
The integral from-1 to 3 of 4x(1-x^2) is -64