What is the integral from 0 to 1 of (x^3-7x-8) ?

The integral from 0 to 1 of (x^3-7x-8) is -45/4

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integral from 0 to 1 of (x^3-7x-8)

\int_{0}^{1} (x^{3} - 7 x - 8) d x

- \frac{45}{4}

Decimal

- 11.25

Solution steps

\int_{0}^{1} x^{3} - 7 x - 8 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_{0}^{1} x^{3} d x - \int_{0}^{1} 7 x d x - \int_{0}^{1} 8 d x

\int_{0}^{1} x^{3} d x = \frac{1}{4}

\int_{0}^{1} 7 x d x = \frac{7}{2}

\int_{0}^{1} 8 d x = 8

= \frac{1}{4} - \frac{7}{2} - 8

Simplify

\frac{1}{4} - \frac{7}{2} - 8 : - \frac{45}{4}

= - \frac{45}{4}

What is the integral from 0 to 1 of (x^3-7x-8) ?
The integral from 0 to 1 of (x^3-7x-8) is -45/4