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

The integral from a to b of (Cx+D)^p is ((D+Cb)^{p+1}-(D+Ca)^{p+1})/(C(p+1))

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integral from a to b of (Cx+D)^p

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

\frac{( D + C b ) ^{p + 1} - ( D + C a ) ^{p + 1}}{C ( p + 1 )}

Solution steps

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

Apply u-substitution

: \int_{D + C a}^{D + C b} \frac{u ^{p}}{C} d u

= \int_{D + C a}^{D + C b} \frac{u ^{p}}{C} d u

Take the constant out:

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

= \frac{1}{C} \cdot \int_{D + C a}^{D + C b} u^{p} d u

Apply the Power Rule

: [\frac{u ^{p + 1}}{p + 1}]_{D + C a}^{D + C b}

= \frac{1}{C} [\frac{u ^{p + 1}}{p + 1}]_{D + C a}^{D + C b}

Simplify

\frac{1}{C} [\frac{u ^{p + 1}}{p + 1}]_{D + C a}^{D + C b} : \frac{[ \frac{u ^{p + 1}}{p + 1} ] _{D + C a}^{D + C b}}{C}

= \frac{[ \frac{u ^{p + 1}}{p + 1} ] _{D + C a}^{D + C b}}{C}

Compute the boundaries:

\frac{( D + C b ) ^{p + 1}}{p + 1} - \frac{( D + C a ) ^{p + 1}}{p + 1}

= \frac{\frac{( D + C b ) ^{p + 1}}{p + 1} - \frac{( D + C a ) ^{p + 1}}{p + 1}}{C}

Simplify

= \frac{( D + C b ) ^{p + 1} - ( D + C a ) ^{p + 1}}{C ( p + 1 )}

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What is the integral from a to b of (Cx+D)^p ?
The integral from a to b of (Cx+D)^p is ((D+Cb)^{p+1}-(D+Ca)^{p+1})/(C(p+1))