What is the integral from-4 to 1 of-x^3-7x^2-8x+16 ?

The integral from-4 to 1 of-x^3-7x^2-8x+16 is 625/12

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integral from-4 to 1 of-x^3-7x^2-8x+16

\int_{- 4}^{1} - x^{3} - 7 x^{2} - 8 x + 16 d x

\frac{625}{12}

Decimal

52.08333 \dots

Solution steps

\int_{- 4}^{1} - x^{3} - 7 x^{2} - 8 x + 16 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}^{1} x^{3} d x - \int_{- 4}^{1} 7 x^{2} d x - \int_{- 4}^{1} 8 x d x + \int_{- 4}^{1} 16 d x

\int_{- 4}^{1} x^{3} d x = - \frac{255}{4}

\int_{- 4}^{1} 7 x^{2} d x = \frac{455}{3}

\int_{- 4}^{1} 8 x d x = - 60

\int_{- 4}^{1} 16 d x = 80

= - (- \frac{255}{4}) - \frac{455}{3} - (- 60) + 80

Simplify

- (- \frac{255}{4}) - \frac{455}{3} - (- 60) + 80 : \frac{625}{12}

= \frac{625}{12}

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What is the integral from-4 to 1 of-x^3-7x^2-8x+16 ?
The integral from-4 to 1 of-x^3-7x^2-8x+16 is 625/12