What is the integral from 1 to 3 of (x-1)^5 ?

The integral from 1 to 3 of (x-1)^5 is 32/3

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integral from 1 to 3 of (x-1)^5

\int_{1}^{3} (x - 1)^{5} d x

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

\int_{1}^{3} (x - 1)^{5} d x

Apply u-substitution

: \int_{0}^{2} u^{5} d u

= \int_{0}^{2} u^{5} d u

Apply the Power Rule

: [\frac{u ^{6}}{6}]_{0}^{2}

= [\frac{u ^{6}}{6}]_{0}^{2}

Compute the boundaries:

\frac{32}{3}

= \frac{32}{3}

\int_{1}^{2} x e^{2 x} d x

\int_{0}^{1} \frac{1}{m x + 1} d x

\int_{x}^{5} \frac{6 ( y - x )}{125} d y

\int_{0}^{2} \frac{2 x}{( x ^{2} + 2 ) ^{2}} d x

\int_{0}^{5} 4 x e^{2 x^{2}} d x

What is the integral from 1 to 3 of (x-1)^5 ?
The integral from 1 to 3 of (x-1)^5 is 32/3