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

The integral from 2 to 1 of (2/(x^2)-3x^2) is 6

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

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

6

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

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

\int_{1}^{2} 3 x^{2} d x = 7

= - (1 - 7)

Simplify

= 6

\int 10 (x - 6)^{4} d x

\int \frac{1}{x ^{2} - x - 2} d x

2 y^{^{''}} + 8 y^{^{'}} = 0

y = x^{2} - 3 x + 2

\int - \frac{3}{2 x} d x

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