What is the integral from 0 to 1 of 19x^3sqrt(1-x^2) ?

The integral from 0 to 1 of 19x^3sqrt(1-x^2) is 38/15

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 19x^3sqrt(1-x^2)

\int_{0}^{1} 19 x^{3} 1 - x^{2} d x

\frac{38}{15}

Decimal

2.53333 \dots

Solution steps

\int_{0}^{1} 19 x^{3} 1 - x^{2} d x

Take the constant out:

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

= 19 \cdot \int_{0}^{1} x^{3} 1 - x^{2} d x

Apply u-substitution

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

= 19 \cdot \int_{1}^{0} - \frac{( - u + 1 ) u}{2} d u

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

= 19 (- \int_{0}^{1} - \frac{( - u + 1 ) u}{2} d u)

Take the constant out:

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

= 19 (- (- \frac{1}{2} \cdot \int_{0}^{1} (- u + 1) u d u))

Expand

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

= 19 (- (- \frac{1}{2} \cdot \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

= 19 (- (- \frac{1}{2} (- \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}

= 19 (- (- \frac{1}{2} (- \frac{2}{5} + \frac{2}{3})))

Simplify

19 (- (- \frac{1}{2} (- \frac{2}{5} + \frac{2}{3}))) : \frac{38}{15}

= \frac{38}{15}

\int_{1}^{2} x^{- 5} d x

\int_{4}^{5} \frac{x}{2 x - 6} d x

\int_{0}^{1} 5 x^{4} + 3 x^{2} + 1 d x

\int_{7}^{8} \frac{1}{x} d x

\int_{1}^{2} \frac{4 ( ln ( x ) ) ^{3}}{x} d x

What is the integral from 0 to 1 of 19x^3sqrt(1-x^2) ?
The integral from 0 to 1 of 19x^3sqrt(1-x^2) is 38/15