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

The integral from 0 to 2 of 3ysqrt(4-y^2) is 8

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 3ysqrt(4-y^2)

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

8

Solution steps

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

Take the constant out:

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

= 3 \cdot \int_{0}^{2} y 4 - y^{2} d y

Apply u-substitution

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

= 3 \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

= 3 (- \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

= 3 (- (- \frac{1}{2} \cdot \int_{0}^{4} u d u))

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{16}{3}

= \frac{3}{2} \cdot \frac{16}{3}

Simplify

= 8

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 3

x \frac{d y}{d x} - 2 y = x^{2}

\frac{d}{d y} (- e^{- x y})

\int \frac{25 - x ^{2}}{x} d x

\int \frac{sec ( x ) tan ( x )}{2 sec ( x ) + 3} d x

What is the integral from 0 to 2 of 3ysqrt(4-y^2) ?
The integral from 0 to 2 of 3ysqrt(4-y^2) is 8