The integral $$int_2^4 1/x^2dx$$ equals:
A. $$1//4$$
B. $$3//4$$
C. $$-3//16$$
D. $$5//16$$

Updated on December 23rd, 2020

First, $\frac{1}{x^{2}}$ can be rewritten as $x^{- 2}$ .

Next, use the integration formula $\int x^{n} d x = \frac{x^{n + 1}}{n + 1}$

$\int x^{- 2} d x = \frac{x^{- 2 + 1}}{- 2 + 1} = \frac{x^{- 1}}{- 1} = - x^{- 1} = - \frac{1}{x}$

Now, calculate at x=4 and x=2 and take the difference between the results.

At x=4: $- \frac{1}{4}$

At x=2: $- \frac{1}{2}$

( $- \frac{1}{4}$ ) - ( $- \frac{1}{2}$ ) = $\frac{1}{4}$

The answer is A.

The integral $$int_2^4 1/x^2dx$$ equals:A. $$1//4$$B. $$3//4$$C. $$-3//16$$D. $$5//16$$