What is the integral from 0 to 1 of 15xsqrt(1-x) ?

The integral from 0 to 1 of 15xsqrt(1-x) is 4

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 15xsqrt(1-x)

\int_{0}^{1} 15 x 1 - x d x

4

Solution steps

\int_{0}^{1} 15 x 1 - x d x

Take the constant out:

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

= 15 \cdot \int_{0}^{1} x 1 - x d x

Apply u-substitution

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

= 15 \cdot \int_{1}^{0} - (- u + 1) u d u

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

= 15 (- \int_{0}^{1} - (- u + 1) u d u)

Take the constant out:

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

= 15 (- (- \int_{0}^{1} (- u + 1) u d u))

Expand

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

= 15 (- (- \int_{0}^{1} - 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

= 15 (- (- (- \int_{0}^{1} u^{\frac{3}{2}} d u + \int_{0}^{1} u d u)))

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

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

= 15 (- (- (- \frac{2}{5} + \frac{2}{3})))

Simplify

15 (- (- (- \frac{2}{5} + \frac{2}{3}))) : 4

= 4

\int_{1}^{6} (2 x + \frac{1}{x}) d x

\int_{x}^{π} sin (t^{2}) d t

\int_{6}^{4} x^{2} d x

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

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

What is the integral from 0 to 1 of 15xsqrt(1-x) ?
The integral from 0 to 1 of 15xsqrt(1-x) is 4