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Popular Calculus >

integral from 0 to 2 of (x^2+4x)

Solution

\int_{0}^{2} (x^{2} + 4 x) d x

Solution

\frac{32}{3}

+1

Decimal

10.66666 \dots

Solution steps

\int_{0}^{2} x^{2} + 4 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_{0}^{2} x^{2} d x + \int_{0}^{2} 4 x d x

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

\int_{0}^{2} 4 x d x = 8

= \frac{8}{3} + 8

Simplify

\frac{8}{3} + 8 : \frac{32}{3}

= \frac{32}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the integral from 0 to 2 of (x^2+4x) ?
The integral from 0 to 2 of (x^2+4x) is 32/3