What is the integral from 0 to t^2 of sqrt(u^7) ?

The integral from 0 to t^2 of sqrt(u^7) is (2t^9)/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to t^2 of sqrt(u^7)

\int_{0}^{t^{2}} u^{7} d u

\frac{2 t ^{9}}{9}

Solution steps

\int_{0}^{t^{2}} u^{7} d u

u^{7} = u^{\frac{7}{2}},

assuming

u \geq 0

= \int_{0}^{t^{2}} u^{\frac{7}{2}} d u

Apply the Power Rule

: [\frac{2}{9} u^{\frac{9}{2}}]_{0}^{t^{2}}

= [\frac{2}{9} u^{\frac{9}{2}}]_{0}^{t^{2}}

Compute the boundaries:

\frac{2}{9} t^{9}

= \frac{2}{9} t^{9}

Simplify

= \frac{2 t ^{9}}{9}

\int_{1}^{2} \frac{( x ^{2} + 2 x + 4 )}{x} d x

\int_{4}^{6} x ln (\frac{x}{6}) d x

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

\int_{- 1}^{0} - x^{3} d x

What is the integral from 0 to t^2 of sqrt(u^7) ?
The integral from 0 to t^2 of sqrt(u^7) is (2t^9)/9