Question

answer the question​

Answer 1

Answer:

6

Explanation:

$\int\limits^3_1 {\frac{9}{x^{2} } } \, dx$

∵9 is constant we can take it out from the integral

$9\int\limits^3_1 {\frac{1}{x^{2} } } \, dx$

We can write this as:

$9\int\limits^3_1 {x^{-2} } \, dx$

We know that:

$\int {x^{n} } \, dx =\frac{x^{n+1} }{n+1}$

∴It is now equal to:

$9[\frac{x^{-2+1} }{-2+1} ]^3_1$

$= 9[\frac{x^{-1} }{-1} ]^3_1$

$=9[-\frac{1}{x} ]^3_1$

$=9[-\frac{1}{3} -(-\frac{1}{1} )]$

$=9[\frac{2}{3}]$

$=6 \;\;\;Answer$