What is the integral from 0 to 4 of (x-6)^3 ?

The integral from 0 to 4 of (x-6)^3 is -320

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integral from 0 to 4 of (x-6)^3

\int_{0}^{4} (x - 6)^{3} d x

- 320

Solution steps

\int_{0}^{4} (x - 6)^{3} d x

Apply u-substitution

: \int_{- 6}^{- 2} u^{3} d u

= \int_{- 6}^{- 2} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{- 6}^{- 2}

= [\frac{u ^{4}}{4}]_{- 6}^{- 2}

Compute the boundaries:

- 320

= - 320

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

\int_{0}^{2} 3 x x^{2} + 4 a^{2} d x

\int_{- 2}^{2} 1 - x^{2} d y

\int_{0}^{1} \frac{1}{2} x^{3} + 2 d x

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

What is the integral from 0 to 4 of (x-6)^3 ?
The integral from 0 to 4 of (x-6)^3 is -320