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

The integral from 0 to 4 of y^2sqrt(4-y) is 2048/105

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{2048}{105}

Decimal

19.50476 \dots

Solution steps

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Expand

(- u + 4)^{2} u : u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 u

= - (- \int_{0}^{4} u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 u 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}^{4} u^{\frac{5}{2}} d u - \int_{0}^{4} 8 u^{\frac{3}{2}} d u + \int_{0}^{4} 16 u d u))

\int_{0}^{4} u^{\frac{5}{2}} d u = \frac{256}{7}

\int_{0}^{4} 8 u^{\frac{3}{2}} d u = \frac{512}{5}

\int_{0}^{4} 16 u d u = \frac{256}{3}

= - (- (\frac{256}{7} - \frac{512}{5} + \frac{256}{3}))

Simplify

- (- (\frac{256}{7} - \frac{512}{5} + \frac{256}{3})) : \frac{2048}{105}

= \frac{2048}{105}

\int_{0}^{9} x - 3 d x

\int_{- 1}^{1} x + 1 - x^{2} d x

\int_{1}^{6} 10 x^{- 4} d x

\int_{1}^{5} (t^{2} - t)^{2} d t

\int_{- 3}^{3} \frac{1}{x} d x

What is the integral from 0 to 4 of y^2sqrt(4-y) ?
The integral from 0 to 4 of y^2sqrt(4-y) is 2048/105