What is the integral from 1 to 2 of (3x^{-2}+x^3) ?

The integral from 1 to 2 of (3x^{-2}+x^3) is 21/4

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integral from 1 to 2 of (3x^{-2}+x^3)

\int_{1}^{2} (3 x^{- 2} + x^{3}) d x

\frac{21}{4}

Decimal Notation

5.25

Solution steps

\int_{1}^{2} 3 x^{- 2} + x^{3} d x

3 x^{- 2} = \frac{3}{x ^{2}}

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

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

\int_{1}^{2} x^{3} d x = \frac{15}{4}

= \frac{3}{2} + \frac{15}{4}

Simplify

\frac{3}{2} + \frac{15}{4} : \frac{21}{4}

= \frac{21}{4}

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What is the integral from 1 to 2 of (3x^{-2}+x^3) ?
The integral from 1 to 2 of (3x^{-2}+x^3) is 21/4