Question

Evaluate : $\int\limits^9_4 {\frac{1}{ \sqrt{x}}} \, dx$

Answer 1

We know that
${(x + a})^{n} = \frac{ {(x + a)}^{n + 1} }{n + 1}$
so here we can write

$\frac{1}{ \sqrt{x} } \: \: as \: \: {x}^{ \frac{ - 1}{2} }$
Now we can apply power rule of Integration

$\int\limits^9_4 {\frac{1}{ \sqrt{x}}} \, dx \\ \\ = \int\limits^9_4 \: {x}^{ - \frac{1}{2} } dx \\ \\ = \frac{ {x}^{ - \frac{1}{2} + 1 } }{ \frac{ - 1}{2} + 1 } \: \: \: \\ \\ \\ = [2\sqrt{x} ]\: \: | 9 - 4 \\ \\ = 2 \sqrt{9} - 2 \sqrt{4} \\ \\ = 2 \times 3 - 2 \times 2 \\ \\ = 6 - 4 \\ \\ \int\limits^9_4 {\frac{1}{ \sqrt{x}}} \, dx = 2$