What is the integral from 0 to 1 of x^2sqrt(x-1) ?

The integral from 0 to 1 of x^2sqrt(x-1) is (16i)/(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 1 of x^2sqrt(x-1)

\int_{0}^{1} x^{2} x - 1 d x

\frac{16 i}{105}

Solution steps

\int_{0}^{1} x^{2} x - 1 d x

Apply u-substitution

: \int_{- 1}^{0} (u + 1)^{2} u d u

= \int_{- 1}^{0} (u + 1)^{2} u d u

Expand

(u + 1)^{2} u : u^{\frac{5}{2}} + 2 u^{\frac{3}{2}} + u

= \int_{- 1}^{0} u^{\frac{5}{2}} + 2 u^{\frac{3}{2}} + 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_{- 1}^{0} u^{\frac{5}{2}} d u + \int_{- 1}^{0} 2 u^{\frac{3}{2}} d u + \int_{- 1}^{0} u d u

\int_{- 1}^{0} u^{\frac{5}{2}} d u = \frac{2 i}{7}

\int_{- 1}^{0} 2 u^{\frac{3}{2}} d u = - \frac{4 i}{5}

\int_{- 1}^{0} u d u = \frac{2 i}{3}

= \frac{2 i}{7} - \frac{4 i}{5} + \frac{2 i}{3}

Simplify

\frac{2 i}{7} - \frac{4 i}{5} + \frac{2 i}{3} : \frac{16 i}{105}

= \frac{16 i}{105}

\int_{0}^{\frac{π}{2}} (e^{t} + cos (t)) d t

\int_{0}^{1} x 2 - x d x

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

\int_{3}^{4} \frac{1}{25 - x ^{2}} d x

\int_{0}^{4} k x d x

What is the integral from 0 to 1 of x^2sqrt(x-1) ?
The integral from 0 to 1 of x^2sqrt(x-1) is (16i)/(105)