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

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

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

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

- \frac{1}{b} + \frac{1}{a}

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

- \frac{1}{b} + \frac{1}{a}

= - \frac{1}{b} + \frac{1}{a}

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

\int \frac{e ^{\frac{x}{2}}}{e ^{x}} d x

(- 1, 2), (1, - 4)

\int 2 x + 4 x^{- 2} d x

x \to 6 lim (9 x + ∣ x - 6 ∣)

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