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

The integral from 0 to 4 of (x^2+4x) is 160/3

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

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

\frac{160}{3}

Decimal

53.33333 \dots

Solution steps

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

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} 4 x d x = 32

= \frac{64}{3} + 32

Simplify

\frac{64}{3} + 32 : \frac{160}{3}

= \frac{160}{3}

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What is the integral from 0 to 4 of (x^2+4x) ?
The integral from 0 to 4 of (x^2+4x) is 160/3