Correct answer: (A) √3. I=∫12x√(4-x^2)dx Let 4-x^2=t^2; xdx=-tdt I=-∫√30t^2dt=-t^33√30=√3

Q116 / 160 · TS EAMCET 2022 (18 Jul Shift 2) · Mathematics

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

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

Solution

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

Hence, the correct answer is option: 'A'.

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∫12x4-x2dx=

Solution

Relevant Questions

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