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Popular Calculus >

integral from-2 to 0 of 8x^2+8x

Solution

\int_{- 2}^{0} 8 x^{2} + 8 x d x

Solution

\frac{16}{3}

+1

Decimal

5.33333 \dots

Solution steps

\int_{- 2}^{0} 8 x^{2} + 8 x 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_{- 2}^{0} 8 x^{2} d x + \int_{- 2}^{0} 8 x d x

\int_{- 2}^{0} 8 x^{2} d x = \frac{64}{3}

\int_{- 2}^{0} 8 x d x = - 16

= \frac{64}{3} - 16

Simplify

\frac{64}{3} - 16 : \frac{16}{3}

= \frac{16}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the integral from-2 to 0 of 8x^2+8x ?
The integral from-2 to 0 of 8x^2+8x is 16/3