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

The integral from 0 to 1 of 2x(2-x^2) is 3/2

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 2x(2-x^2)

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

\frac{3}{2}

Decimal Notation

1.5

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

: [\frac{1}{2} x^{2} (2 - x^{2}) - \int - x^{3} d x]_{0}^{1}

= 2 [\frac{1}{2} x^{2} (2 - x^{2}) - \int - x^{3} d x]_{0}^{1}

\int - x^{3} d x = - \frac{x ^{4}}{4}

= 2 [\frac{1}{2} x^{2} (2 - x^{2}) - (- \frac{x ^{4}}{4})]_{0}^{1}

Simplify

2 [\frac{1}{2} x^{2} (2 - x^{2}) - (- \frac{x ^{4}}{4})]_{0}^{1} : 2 [\frac{1}{4} (4 x^{2} - x^{4})]_{0}^{1}

= 2 [\frac{1}{4} (4 x^{2} - x^{4})]_{0}^{1}

Compute the boundaries:

\frac{3}{4}

= 2 \cdot \frac{3}{4}

Simplify

= \frac{3}{2}

\int_{0}^{1} (5 x^{e} + 4 e^{x}) d x

\int_{0}^{3} (x^{2} - 5 x + 9) d x

\int_{2}^{4} x^{2} + 5 x + 7 + 1 d x

\int_{- 7}^{4} \frac{1}{2} x d x

What is the integral from 0 to 1 of 2x(2-x^2) ?
The integral from 0 to 1 of 2x(2-x^2) is 3/2