-0 a x dx -22- x 2-A- a -2-1B- a 1-2C- a 1-2D- 2 a -3

The correct option is A a(√(2) 1)Put t=a^2+x^2 ⇒ 2 xdx=dt, then ∫_0^a x dx/√(a^2+x^2)=1/2∫_a^2^2a^21/√(t)dt =[(2a^2)^1/2 a^2/2]=a(√(2) 1) ...

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Byju's Answer

Standard XII

Mathematics

Property 1

∫0 a x dx /√2...

Question

$\int_{0}^{a} \frac{x d x}{\sqrt{a^{2} + x^{2}}} =$

a (\sqrt{2} - 1)

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a (1 - \sqrt{2})

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a (1 + \sqrt{2})

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2 a \sqrt{3}

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Solution

The correct option is A $a (\sqrt{2} - 1)$
$P u t t = a^{2} + x^{2} \Rightarrow 2 x d x = d t, t h e n$
$\int_{0}^{a} \frac{x d x}{\sqrt{a^{2} + x^{2}}} = \frac{1}{2} \int_{a^{2}}^{2 a^{2}} \frac{1}{\sqrt{t}} d t$
$= [(2 a^{2})^{\frac{1}{2}} - a^{\frac{2}{2}}] = a (\sqrt{2} - 1)$

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∫a0x dx√a2+x2=

The correct option is A a(√2−1)Put t=a2+x2⇒2 xdx=dt, then ∫a0x dx√a2+x2=12∫2a2a21√tdt =[(2a2)12−a22]=a(√2−1)

$\int_{0}^{a} \frac{x d x}{\sqrt{a^{2} + x^{2}}} =$

The correct option is A $a (\sqrt{2} - 1)$
$P u t t = a^{2} + x^{2} \Rightarrow 2 x d x = d t, t h e n$
$\int_{0}^{a} \frac{x d x}{\sqrt{a^{2} + x^{2}}} = \frac{1}{2} \int_{a^{2}}^{2 a^{2}} \frac{1}{\sqrt{t}} d t$
$= [(2 a^{2})^{\frac{1}{2}} - a^{\frac{2}{2}}] = a (\sqrt{2} - 1)$