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

The integral from 1/4 to 1 of sqrt(x)-x^2 is 49/192

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

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

\frac{49}{192}

Solution steps

\int_{\frac{1}{4}}^{1} x - x^{2} 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_{\frac{1}{4}}^{1} x d x - \int_{\frac{1}{4}}^{1} x^{2} d x

\int_{\frac{1}{4}}^{1} x d x = \frac{7}{12}

\int_{\frac{1}{4}}^{1} x^{2} d x = \frac{21}{64}

= \frac{7}{12} - \frac{21}{64}

Simplify

\frac{7}{12} - \frac{21}{64} : \frac{49}{192}

= \frac{49}{192}

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What is the integral from 1/4 to 1 of sqrt(x)-x^2 ?
The integral from 1/4 to 1 of sqrt(x)-x^2 is 49/192