What is the integral from 0 to 1 of xsqrt(ax^2+b) ?

The integral from 0 to 1 of xsqrt(ax^2+b) is ((a+b)^{3/2}-b^{3/2})/(3a)

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 xsqrt(ax^2+b)

\int_{0}^{1} x a x^{2} + b d x

\frac{( a + b ) ^{\frac{3}{2}} - b ^{\frac{3}{2}}}{3 a}

Solution steps

\int_{0}^{1} x a x^{2} + b d x

Apply u-substitution

: \int_{b}^{a + b} \frac{u}{2 a} d u

= \int_{b}^{a + b} \frac{u}{2 a} d u

Take the constant out:

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

= \frac{1}{2 a} \cdot \int_{b}^{a + b} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{b}^{a + b}

= \frac{1}{2 a} [\frac{2}{3} u^{\frac{3}{2}}]_{b}^{a + b}

Simplify

\frac{1}{2 a} [\frac{2}{3} u^{\frac{3}{2}}]_{b}^{a + b} : \frac{[ \frac{2}{3} u ^{\frac{3}{2}} ] _{b}^{a + b}}{2 a}

= \frac{[ \frac{2}{3} u ^{\frac{3}{2}} ] _{b}^{a + b}}{2 a}

Compute the boundaries:

\frac{2}{3} (a + b)^{\frac{3}{2}} - \frac{2}{3} b^{\frac{3}{2}}

= \frac{\frac{2}{3} ( a + b ) ^{\frac{3}{2}} - \frac{2}{3} b ^{\frac{3}{2}}}{2 a}

Simplify

= \frac{( a + b ) ^{\frac{3}{2}} - b ^{\frac{3}{2}}}{3 a}

sin^{2} (θ)

y^{^{'}} = \frac{6 t ^{2}}{y}

\int_{- 1}^{1} x^{2} 5 - 3 x d x

\frac{3}{s ^{2} + 2 s + 5}

\frac{d}{d x} (2 x^{4} + 3 x - 3)

What is the integral from 0 to 1 of xsqrt(ax^2+b) ?
The integral from 0 to 1 of xsqrt(ax^2+b) is ((a+b)^{3/2}-b^{3/2})/(3a)