What is the integral from 0 to 1 of x^2(4-x^2) ?

The integral from 0 to 1 of x^2(4-x^2) is 17/15

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integral from 0 to 1 of x^2(4-x^2)

\int_{0}^{1} x^{2} (4 - x^{2}) d x

\frac{17}{15}

Decimal

1.13333 \dots

Solution steps

\int_{0}^{1} x^{2} (4 - x^{2}) d x

Expand

x^{2} (4 - x^{2}) : 4 x^{2} - x^{4}

= \int_{0}^{1} 4 x^{2} - x^{4} 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} 4 x^{2} d x - \int_{0}^{1} x^{4} d x

\int_{0}^{1} 4 x^{2} d x = \frac{4}{3}

\int_{0}^{1} x^{4} d x = \frac{1}{5}

= \frac{4}{3} - \frac{1}{5}

Simplify

\frac{4}{3} - \frac{1}{5} : \frac{17}{15}

= \frac{17}{15}

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What is the integral from 0 to 1 of x^2(4-x^2) ?
The integral from 0 to 1 of x^2(4-x^2) is 17/15