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

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

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

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

- \frac{1}{2}

Decimal

- 0.5

Solution steps

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

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

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

Apply the Power Rule

: [- \frac{1}{x}]_{1}^{2}

= - [- \frac{1}{x}]_{1}^{2}

Compute the boundaries:

\frac{1}{2}

= - \frac{1}{2}

\int_{0}^{3} \frac{x}{1 + x} d x

\int_{2}^{3} (x^{2} + x - 12) d x

\int_{0}^{π} sin (x) sin (6 x) d x

\int_{0}^{\frac{π}{3}} (5 sec^{2} (x)) d x

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

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