What is the integral from a to b of (ax+b)^p ?

The integral from a to b of (ax+b)^p is ((b+ab)^{p+1}-(b+a^2)^{p+1})/(a(p+1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from a to b of (ax+b)^p

\int_{a}^{b} (a x + b)^{p} d x

\frac{( b + ab ) ^{p + 1} - ( b + a ^{2} ) ^{p + 1}}{a ( p + 1 )}

Solution steps

\int_{a}^{b} (a x + b)^{p} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{( b + ab ) ^{p + 1}}{p + 1} - \frac{( b + a ^{2} ) ^{p + 1}}{p + 1}

= \frac{\frac{( b + ab ) ^{p + 1}}{p + 1} - \frac{( b + a ^{2} ) ^{p + 1}}{p + 1}}{a}

Simplify

= \frac{( b + ab ) ^{p + 1} - ( b + a ^{2} ) ^{p + 1}}{a ( p + 1 )}

\int_{0}^{11} 121 - x^{2} d x

\int_{0}^{\frac{π}{2}} \frac{1 - cos ( 2 x )}{2} d x

\int_{0}^{\frac{π}{4}} sin^{5} (x) d x

\int_{- 4}^{0} x^{2} + 4 x d x

\int_{- 1}^{1} x^{2} sin (nπ x) d x

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