What is the integral from 0 to 4 of (x^2+9x) ?

The integral from 0 to 4 of (x^2+9x) is 280/3

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integral from 0 to 4 of (x^2+9x)

\int_{0}^{4} (x^{2} + 9 x) d x

\frac{280}{3}

Decimal Notation

93.33333 \dots

Solution steps

\int_{0}^{4} x^{2} + 9 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} 9 x d x

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

\int_{0}^{4} 9 x d x = 72

= \frac{64}{3} + 72

Simplify

\frac{64}{3} + 72 : \frac{280}{3}

= \frac{280}{3}

\int_{0}^{3} x \frac{x ^{2}}{9} d x

\int_{2}^{3} \frac{1}{x ^{4}} d x

\int_{2}^{4} x^{2} e^{x^{3} + 5} d x

\int_{- 1}^{5} (4 x^{3} - 3 x^{2} + 1) x d x

\int_{2}^{y} x^{2} sin (x) d x

What is the integral from 0 to 4 of (x^2+9x) ?
The integral from 0 to 4 of (x^2+9x) is 280/3