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

The integral from 1 to 4 of sqrt(x)-1/(x^2) is 47/12

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

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

\frac{47}{12}

Solution steps

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

\int_{1}^{4} x d x = \frac{14}{3}

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

= \frac{14}{3} - \frac{3}{4}

Simplify

\frac{14}{3} - \frac{3}{4} : \frac{47}{12}

= \frac{47}{12}

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What is the integral from 1 to 4 of sqrt(x)-1/(x^2) ?
The integral from 1 to 4 of sqrt(x)-1/(x^2) is 47/12