Question

integral 1 to 9(( x-1)÷√x)​

Answer 1

Step-by-step explanation:

We have,

$\int \limits^{9}_{1} \frac{x - 1}{ \sqrt{x} } dx$

$= \int \limits^{9}_{1} ( \sqrt{x} - \frac{1 }{ \sqrt{x} })dx$

$= \int \limits^{9}_{1} \sqrt{x}dx - \int \limits^{9}_{1} \frac{1}{ \sqrt{x} } dx$

$= \int \limits^{9}_{1} {x}^{ \frac{1}{2} } dx - \int \limits^{9}_{1} {x}^{ - \frac{1}{2} } dx$

$= [ \frac{2}{3} {x}^{ \frac{3}{2} } ]^{9}_{1} - [2 \sqrt{x} ]^{9}_{1}$

$= \frac{2}{3} ({9})^{ \frac{3}{2} } - \frac{2}{3} ({1})^{ \frac{3}{2} } - 2 \sqrt{9} + 2 \sqrt{1}$

$= \frac{2}{3} \times(3)^{3} - \frac{2}{3} - 2 \times 3 + 2$

$= 2 \times ( {3})^{2} - \frac{2}{3} - 6 + 2$

$= 18 - 6 + 2 - \frac{2}{3}$

$= 14 - \frac{2}{3}$

$= \frac{40}{3}$