Evaluate - 0 1 - x 1- x 3 dx-

The correct option is B π2 ∫^1_0 √(x1 x^3)dx put #160; t=x^32 dt=32x^12dx when x=1,t=1 and x=0,t=0 ∫^1_0 dt√(1 t^2)=[sin^ 1t]^ ...

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Byju's Answer

Standard XII

Physics

Non Uniform Motion & Avg Velocity

Evaluate ∫ ...

Question

Evaluate $\int_{0}^{1} \sqrt{\frac{x}{1 - x^{3}}} d x =$

\frac{π}{4}

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\frac{π}{3}

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\frac{π}{6}

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\frac{π}{2}

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Solution

The correct option is B $\frac{π}{2}$
$\int_{0}^{1} \sqrt{\frac{x}{1 - x^{3}}} d x$
put $t = x^{\frac{3}{2}} ⟹ d t = \frac{3}{2} x^{\frac{1}{2}} d x$
when $x = 1, t = 1$ and $x = 0, t = 0$
$\int_{0}^{1} \frac{d t}{\sqrt{1 - t^{2}}} = [{sin}^{- 1} t]_{0}^{1} = \frac{π}{2}$

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Evaluate ∫10√x1−x3dx=

The correct option is B π2∫10√x1−x3dxput t=x32⟹dt=32x12dxwhen x=1,t=1 and x=0,t=0∫10dt√1−t2=[sin−1t]10=π2

Evaluate $\int_{0}^{1} \sqrt{\frac{x}{1 - x^{3}}} d x =$

The correct option is B $\frac{π}{2}$
$\int_{0}^{1} \sqrt{\frac{x}{1 - x^{3}}} d x$
put $t = x^{\frac{3}{2}} ⟹ d t = \frac{3}{2} x^{\frac{1}{2}} d x$
when $x = 1, t = 1$ and $x = 0, t = 0$
$\int_{0}^{1} \frac{d t}{\sqrt{1 - t^{2}}} = [{sin}^{- 1} t]_{0}^{1} = \frac{π}{2}$