Question

${\displaystyle \sf \bold \red{\int^{2}_{ - 2} \bigg( {x}^{3} \cos \frac{x}{2} + \frac{1}{2} \bigg) \sqrt{4 - {x}^{2} } \: dx}}$​

Answer 1

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

$\displaystyle \sf \red{ = \int^{2}_{ - 2} \bigg( {x}^{3} \cos \frac{x}{2} \frac{}{} \bigg) \sqrt{4 - {x}^{2} } \: dx + \int^{2}_{ - 2} \bigg( \frac{1}{2} \bigg) \sqrt{4 - {x}^{2} } \: dx}$

$\sf \red{ \bigg({ {x}^{3} }^{} cos \frac{x}{2} \bigg) \sqrt{4 - {x}^{2} } \: is \: an \: odd \: function}$

Recall the integral of an odd function over a symmetric intevral [-a, a] is zero.

$\displaystyle\sf \red{\implies \int^{2}_{ - 2} \bigg( {x}^{3} cos \frac{x}{2} \bigg ) \sqrt{4 - {x}^{2} }\: dx = 0}$

$\displaystyle\sf \red{ = \int^{2}_{ - 2} \bigg( {}^{} \frac{1}{2} \bigg) \sqrt{4 - {x}^{2} } \: dx }$

x²+y²=4 is a circle centered at (0,0) with a radius of 2.

$\displaystyle \sf \red{ \implies \int^{2}_{ - 2} \bigg( \frac{1}{2} \bigg) \sqrt{4 - {x}^{2} } \: dx = half \: the \: area \: of \: (semi \: circle , \: radius \: 2)}$

$\sf \red{ = \bigg( \frac{1}{2} \bigg ) \bigg( \frac{1}{2} \bigg) \pi(2) {}^{2} = \pi}$