What is the integral from 0 to 4 of (x^2+x-6) ?

The integral from 0 to 4 of (x^2+x-6) is 16/3

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integral from 0 to 4 of (x^2+x-6)

\int_{0}^{4} (x^{2} + x - 6) d x

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

\int_{0}^{4} x^{2} + x - 6 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} x d x - \int_{0}^{4} 6 d x

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

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

\int_{0}^{4} 6 d x = 24

= \frac{64}{3} + 8 - 24

Simplify

\frac{64}{3} + 8 - 24 : \frac{16}{3}

= \frac{16}{3}

\int_{0}^{2} x^{2} - 4 x + 3 d x

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What is the integral from 0 to 4 of (x^2+x-6) ?
The integral from 0 to 4 of (x^2+x-6) is 16/3