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

The integral from 0 to 1 of (sqrt(x)-x^2)^2 is 9/70

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

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

\frac{9}{70}

Decimal Notation

0.12857 \dots

Solution steps

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

Expand

(x - x^{2})^{2} : x - 2 x^{\frac{5}{2}} + x^{4}

= \int_{0}^{1} x - 2 x^{\frac{5}{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} x d x - \int_{0}^{1} 2 x^{\frac{5}{2}} d x + \int_{0}^{1} x^{4} d x

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

\int_{0}^{1} 2 x^{\frac{5}{2}} d x = \frac{4}{7}

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

= \frac{1}{2} - \frac{4}{7} + \frac{1}{5}

Simplify

\frac{1}{2} - \frac{4}{7} + \frac{1}{5} : \frac{9}{70}

= \frac{9}{70}

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What is the integral from 0 to 1 of (sqrt(x)-x^2)^2 ?
The integral from 0 to 1 of (sqrt(x)-x^2)^2 is 9/70