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

The integral from 0 to 2 of 2xsqrt(4-x^2) 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 0 to 2 of 2xsqrt(4-x^2)

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

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{0} - \frac{u}{2} d u

= 2 \cdot \int_{4}^{0} - \frac{u}{2} d u

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

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

Take the constant out:

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

= 2 (- (- \frac{1}{2} \cdot \int_{0}^{4} u d u))

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

= 2 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}))

Simplify

2 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4})) : [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

Compute the boundaries:

\frac{16}{3}

= \frac{16}{3}

\int_{1}^{e} ln (y) d y

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

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

\int_{- 1}^{0} - 5 x (x^{2} - 1)^{4} d x

\int_{0}^{1} (x + 8 1 - x^{2}) d x

What is the integral from 0 to 2 of 2xsqrt(4-x^2) ?
The integral from 0 to 2 of 2xsqrt(4-x^2) is 16/3