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//41//4B. 3//43//4C. -3//16-3//16D. 5//165//16