What is the integral from 1 to 3 of (x-3)^4 ?

The integral from 1 to 3 of (x-3)^4 is 32/5

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integral from 1 to 3 of (x-3)^4

\int_{1}^{3} (x - 3)^{4} d x

\frac{32}{5}

Decimal Notation

6.4

Solution steps

\int_{1}^{3} (x - 3)^{4} d x

Apply u-substitution

: \int_{- 2}^{0} u^{4} d u

= \int_{- 2}^{0} u^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{- 2}^{0}

= [\frac{u ^{5}}{5}]_{- 2}^{0}

Compute the boundaries:

\frac{32}{5}

= \frac{32}{5}

\int_{0}^{4} (x)^{2} d x

\int_{0}^{5} \frac{x}{49 + 3 x} d x

\int_{0}^{0.6} sin (x^{2}) d x

\int_{\frac{π}{6}}^{\frac{π}{3}} cos (x) d x

\int_{0}^{y} 8 x y d x

What is the integral from 1 to 3 of (x-3)^4 ?
The integral from 1 to 3 of (x-3)^4 is 32/5