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

The integral from 1 to 2 of (3-6/(x^2)) is 0

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

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

0

Solution steps

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

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

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

= 3 - 3

Simplify

= 0

f (x) = (1 - 2 x) (1 + 2 x), at x = 2

n = 1 \sum \infty (\frac{3}{8})^{n - 1}

y = \frac{x ^{2} - 2}{2 x + 1}

\int_{0}^{2} x - x^{2} d x

\int sin^{5} (θ) cos (θ) d θ

What is the integral from 1 to 2 of (3-6/(x^2)) ?
The integral from 1 to 2 of (3-6/(x^2)) is 0