Question

Solve
$\bf\displaystyle \int^4_2\frac{\sqrt{x} }{\sqrt{6 - x + x} } dx$

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Given:

$\bf\displaystyle \int^4_2\frac{\sqrt{x} }{\sqrt{6 - x + x} } dx$

To find : Integral

Explanation:

$\bf\displaystyle \int^4_2\frac{\sqrt{x} }{\sqrt{6 } } dx$

$\bf\displaystyle \frac{1}{\sqrt{6 } }\int^4_2\sqrt{x} dx$

Do integration

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

$=\bf\displaystyle\frac{2}{\sqrt{6} } [\frac{x^{\frac{3}{2}}}{3}]$

put lower and upper limits

$= \frac{2}{ \sqrt{6} } ( \frac{4 \times 2}{3} - \frac{2 \sqrt{2} }{3} ) \\ \\ = \frac{2}{3 \sqrt{6} } (8 - 2\sqrt{2} )$

or

$= \frac{2}{ \sqrt{6} } ( \frac{4 \times 2}{3} - \frac{2 \sqrt{2} }{3} ) \\ \\ = \frac{4}{3 \sqrt{6} } (4 - \sqrt{2} )$

Thus,

$\bf\displaystyle \int^4_2\frac{\sqrt{x} }{\sqrt{6 - x + x} } dx=\frac{4}{3 \sqrt{6} } (4 - \sqrt{2} )$

Hope it helps you.