What is the integral from 0 to 2 of x^3sqrt((4-x^2)) ?

The integral from 0 to 2 of x^3sqrt((4-x^2)) is 64/15

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 2 of x^3sqrt((4-x^2))

\int_{0}^{2} x^{3} (4 - x^{2}) d x

\frac{64}{15}

Decimal

4.26666 \dots

Solution steps

\int_{0}^{2} x^{3} 4 - x^{2} d x

Apply u-substitution

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

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

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

Expand

(- u^{2} + 4) u^{2} : - u^{4} + 4 u^{2}

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

Apply the Sum Rule:

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

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

\int_{0}^{2} u^{4} d u = \frac{32}{5}

\int_{0}^{2} 4 u^{2} d u = \frac{32}{3}

= - (- (- \frac{32}{5} + \frac{32}{3}))

Simplify

- (- (- \frac{32}{5} + \frac{32}{3})) : \frac{64}{15}

= \frac{64}{15}

\int_{2}^{10} e^{x} d x

\int_{0}^{\frac{7}{3}} \frac{1}{9 x ^{2} + 49} d x

\int_{- 3}^{3} 81 - x^{4} d x

\int_{0}^{0} cos (θ) d θ

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

What is the integral from 0 to 2 of x^3sqrt((4-x^2)) ?
The integral from 0 to 2 of x^3sqrt((4-x^2)) is 64/15