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

The integral from 1 to 3 of (x^3+2)/(x^2) is 16/3

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integral from 1 to 3 of (x^3+2)/(x^2)

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

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

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

Expand

\frac{x ^{3} + 2}{x ^{2}} : x + \frac{2}{x ^{2}}

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

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

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

= 4 + \frac{4}{3}

Simplify

4 + \frac{4}{3} : \frac{16}{3}

= \frac{16}{3}

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What is the integral from 1 to 3 of (x^3+2)/(x^2) ?
The integral from 1 to 3 of (x^3+2)/(x^2) is 16/3