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

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

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

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

\frac{1}{2}

Decimal

0.5

Solution steps

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

Expand

\frac{x ^{2} - 1}{x ^{2}} : 1 - \frac{1}{x ^{2}}

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

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

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

= 1 - \frac{1}{2}

Simplify

1 - \frac{1}{2} : \frac{1}{2}

= \frac{1}{2}

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

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

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