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

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

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integral from 4 to 5 of (x-4)-(x-4)^3

\int_{4}^{5} (x - 4) - (x - 4)^{3} d x

\frac{1}{4}

Decimal

0.25

Solution steps

\int_{4}^{5} x - 4 - (x - 4)^{3} 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_{4}^{5} x d x - \int_{4}^{5} 4 d x - \int_{4}^{5} (x - 4)^{3} d x

\int_{4}^{5} x d x = \frac{9}{2}

\int_{4}^{5} 4 d x = 4

\int_{4}^{5} (x - 4)^{3} d x = \frac{1}{4}

= \frac{9}{2} - 4 - \frac{1}{4}

Simplify

\frac{9}{2} - 4 - \frac{1}{4} : \frac{1}{4}

= \frac{1}{4}

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