Question

2
( x3 cos 2x + 2 1 √4-x2 dx


-2​

Answer 1

Step-by-step explanation:

We have,

$\int _{ - 2}^{2} ( {x}^{3} \cos( \frac{x}{2} ) + \frac{1}{2} ) \sqrt{4 - {x}^{2} } dx$

$= \int _{ - 2}^{2} {x}^{3} \cos( \frac{x}{2} ) \sqrt{4 - {x}^{2} } dx - \int _{ - 2}^{2} \frac{1}{2} \sqrt{4 - {x}^{2} } dx$

$since \: \: {x}^{3} \cos( \frac{x}{2} ) \sqrt{4 - {x}^{2} } \: \: is \: \: a \: \: odd \: \: function$

so,

$\int _{ - 2}^{2} {x}^{3} \cos( \frac{x}{2} ) \sqrt{4 - {x}^{2} } dx = 0$

so, required integral is

$= - \frac{1}{2} \int _{ - 2}^{2} \sqrt{4 - {x}^{2} } dx$

$= - \frac{1}{2} \times 2\int _{ 0}^{2} \sqrt{ {(2)}^{2} - {(x)}^{2} } dx$

$= - [ \frac{x}{2} \sqrt{4 - {x}^{2} } ] _{ 0}^{2} - [ \frac{4 }{2} \sin^{ - 1} ( \frac{x}{2} ) ]_{0}^{2}$

$= - [ \frac{2}{2} \sqrt{4 - {2}^{2} } ] - [ \frac{4 }{2} \sin^{ - 1} ( \frac{2}{2} ) ] + 0$

$= - {\pi}$