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

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

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

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

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

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

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

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

\frac{\partial}{\partial x} (e^{x} sin (2 y))

\int \frac{2 x + 4}{x} d x

x \to - 3 lim (\frac{( x + 3 )}{x - 3})

\int 5 (3 - 4 x)^{3} d x

x \to 8 lim (3 x - 2)

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