What is the integral from 1 to 2 of (8x^3+6/(x^2)) ?

The integral from 1 to 2 of (8x^3+6/(x^2)) is 33

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integral from 1 to 2 of (8x^3+6/(x^2))

\int_{1}^{2} (8 x^{3} + \frac{6}{x ^{2}}) d x

33

Solution steps

\int_{1}^{2} 8 x^{3} + \frac{6}{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}^{2} 8 x^{3} d x + \int_{1}^{2} \frac{6}{x ^{2}} d x

\int_{1}^{2} 8 x^{3} d x = 30

\int_{1}^{2} \frac{6}{x ^{2}} d x = 3

= 30 + 3

Simplify

= 33

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What is the integral from 1 to 2 of (8x^3+6/(x^2)) ?
The integral from 1 to 2 of (8x^3+6/(x^2)) is 33