∫ 1 2 x 4 - x 2 d x =, 3, 2, 1 3, 1 2

I = ∫ 1 2 x 4 - x 2 d x Let 4 - x 2 = t 2 ; x d x = - t d t I = - ∫ 3 0 t 2 d t = - t 3 3 3 0 = 3

∫ 1 2 x 4 - x 2 d x =

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Question: Answered & Verified by Expert

\int_{1}^{2} x \sqrt{4 - x^{2}} d x =

MathematicsDefinite IntegrationTS EAMCETTS EAMCET 2022 (18 Jul Shift 2)

Options:

A $\sqrt{3}$
B $2$
C $\frac{1}{\sqrt{3}}$
D $\frac{1}{2}$

Solution:

2672 Upvotes Verified Answer

The correct answer is:

\sqrt{3}

$I = \int_{1}^{2} x \sqrt{4 - x^{2}} d x$
Let $4 - x^{2} = t^{2}; x d x = - t d t$
$I = - \int_{\sqrt{3}}^{0} t^{2} d t = - {[\frac{t^{3}}{3}]}_{\sqrt{3}}^{0} = \sqrt{3}$

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