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BITSAT 2018 Paper

Section: Mathematics

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Question : 112 of 150

Marks: +1, -0

The value of the integral

\int\limits_{a}^{b}

\frac{\sqrt{x}\,dx}{\sqrt{x}+\sqrt{a+b-x}}

is :

π
$\frac{1}{2}$ (b - a)
π/2
b - a

Solution:

Given, I =

\int\limits_{a}^{b}

\frac{\sqrt{x}\,dx}{\sqrt{x}+\sqrt{a+b-x}}

... (i)

Note :

\int\limits_{a}^{b}

f (x) dx =

\int\limits_{a}^{b}

f (a + b - x) dx

∴ i =

\int\limits_{a}^{b}

\frac{\sqrt{a+b-x}}{\sqrt{a+b-x}+\sqrt{x}}

... (ii)

Add (i) and (ii),

2I =

\int\limits_{a}^{b}

\frac{\sqrt{x}\,dx}{\sqrt{x}+\sqrt{a+b-x}}

+

\int\limits_{a}^{b}

\frac{\sqrt{a+b-x}\,dx}{\sqrt{a+b-x}+\sqrt{x}}

=

\int\limits_{a}^{b}

\frac{\sqrt{x}+\sqrt{a+b-x}\,dx}{\sqrt{a+b+x}+\sqrt{x}}

=

\int\limits_{a}^{b}

1 . dx =

\left[x\right]_{a}^{b}

2I = b – a

∴ I =

\frac{b-a}{2}

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