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

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

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

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

\frac{1}{4}

Decimal Notation

0.25

Solution steps

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

Simplify

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int_{3}^{4} x d x = \frac{7}{2}

\int_{3}^{4} 4 d x = 4

= - \frac{1}{4} - \frac{7}{2} + 4

Simplify

- \frac{1}{4} - \frac{7}{2} + 4 : \frac{1}{4}

= \frac{1}{4}

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