What is the integral from 0 to 1 of (9x^2+18x^3) ?

The integral from 0 to 1 of (9x^2+18x^3) is 15/2

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integral from 0 to 1 of (9x^2+18x^3)

\int_{0}^{1} (9 x^{2} + 18 x^{3}) d x

\frac{15}{2}

Decimal

7.5

Solution steps

\int_{0}^{1} 9 x^{2} + 18 x^{3} 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}^{1} 9 x^{2} d x + \int_{0}^{1} 18 x^{3} d x

\int_{0}^{1} 9 x^{2} d x = 3

\int_{0}^{1} 18 x^{3} d x = \frac{9}{2}

= 3 + \frac{9}{2}

Simplify

3 + \frac{9}{2} : \frac{15}{2}

= \frac{15}{2}

What is the integral from 0 to 1 of (9x^2+18x^3) ?
The integral from 0 to 1 of (9x^2+18x^3) is 15/2