Please solve it..... Don't spam

1.	Please solve it..... Don't spam
Answer» GIVEN :– $\\ \: \: { \bold{I= \int^<klux>2</klux> _{-2} \left( {x}^{3} \cos \dfrac{x}{2} + \dfrac{<klux>1</klux>}{2} \right) \sqrt{4 - {x}^{2} }.dx }}\\$ TO FIND :– • Value of 'I' = ? SOLUTION :– $\\ \: \: { \bold{I= \int^2 _{-2} \left( {x}^{3} \cos \dfrac{x}{2} + \dfrac{1}{2} \right) \sqrt{4 - {x}^{2} }.dx \:\:\:----eq.(1)}}\\$ • We know that – $\\ \: \: \dashrightarrow \: \: { \bold{ \int^b _{a} f(x).dx =\int^b _{a} f(a + b - x).dx }}\\$ • So that – $\\ \implies \: \: { \bold{I= \int^2 _{-2} \left( {( - x)}^{3} \cos \dfrac{ - x}{2} + \dfrac{1}{2} \right) \sqrt{4 - {( - x)}^{2} }.dx }}\\$ $\\ \implies \: \: { \bold{I= \int^2 _{-2} \left( -{ x}^{3} \cos \dfrac{x}{2} + \dfrac{1}{2} \right) \sqrt{4 - {x}^{2} }.dx \:\:\:----eq.(2)}}\\$ • Now ADD eq.(1) & eq.(2) – $\\ \implies \: \: { \bold{2I= \int^2 _{-2} \left \{ \left( -{ x}^{3} \cos \dfrac{x}{2} + \dfrac{1}{2} \right) \sqrt{4 - {x}^{2} } + \left( {x}^{3} \cos \dfrac{x}{2} + \dfrac{1}{2} \right) \sqrt{4 - {x}^{2} }\right\}.dx}}\\$ $\\ \implies \: \: { \bold{2I= \int^2 _{-2} \left( \dfrac{1}{2} + \dfrac{1}{2} \right) \sqrt{4 - {x}^{2} } .dx}}\\$ $\\ \implies \: \: { \bold{2I= \int^2 _{-2} \sqrt{4 - {x}^{2} } .dx}}\\$ • Using PROPERTY – $\\ \dashrightarrow \: \: { \bold{\int^a _{-a} f(x).dx =2\int^a _{0} f(x).dx \: \: \: \: \: \:[ f(x) = even \: \: function]}}\\$ • So – $\\ \implies \: \: { \bold{2I= 2\int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\$ $\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\$ • Now put x = 2 $\sin( \theta)$ – $\\ \implies \: \: { \bold{x = 2 \sin( \theta)}} \\$ $\\ \implies \: \: { \bold{dx = 2 \cos( \theta).d \theta}} \\$ $\\ \implies \: \: { \bold{I= \int^ \frac{\pi}{2} _{0} \sqrt{4 - {(2 \sin \theta) }^{2} }(2 \cos\theta).d \theta}}\\$ $\\ \implies \: \: { \bold{I= \int^ \frac{\pi}{2} _{0} \sqrt{4 - {4 \sin^{2} \theta} }(2 \cos\theta).d \theta}}\\$ $\\ \implies \: \: { \bold{I= \int^ \frac{\pi}{2} _{0} 2 \sqrt{1 - {\sin^{2} \theta} }(2 \cos\theta).d \theta}}\\$ $\\ \implies \: \: { \bold{I= 2 \int^ \frac{\pi}{2} _{0} 2\cos^{2} \theta.d \theta}}\\$ $\\ \implies \: \: { \bold{I= 2 \int^ \frac{\pi}{2} _{0} 1 + \cos(2 \theta).d \theta}}\\$ $\\ \implies \: \: { \bold{I= 2 \left[ \theta+ \dfrac{ \sin(2 \theta)}{2} \right]^ \frac{\pi}{2} _{0} }}\\$ $\\ \implies \: \: { \bold{I= 2 \left[ \left( \dfrac{\pi}{2} - 0 \right)+ \dfrac{ \sin(2 \times \dfrac{\pi}{2})}{2} - \sin(0) \right]}}\\$ $\\ \implies \: \: { \bold{I= 2 \left[ \left( \dfrac{\pi}{2} \right)+ \dfrac{ \sin( \pi)}{2} - 0 \right]}}\\$ $\\ \implies \: \: { \bold{I= 2 \left[ \dfrac{\pi}{2} \right]}}\\$ $\\ \implies \large \: \:{ \boxed { \bold{I= \pi}}}\\$

Answer»

GIVEN :–

$\\ \: \: { \bold{I= \int^<klux>2</klux> _{-2} \left( {x}^{3} \cos \dfrac{x}{2} + \dfrac{<klux>1</klux>}{2} \right) \sqrt{4 - {x}^{2} }.dx }}\\$

TO FIND :–

• Value of 'I' = ?

SOLUTION :–

$\\ \: \: { \bold{I= \int^2 _{-2} \left( {x}^{3} \cos \dfrac{x}{2} + \dfrac{1}{2} \right) \sqrt{4 - {x}^{2} }.dx \:\:\:----eq.(1)}}\\$

• We know that –

$\\ \: \: \dashrightarrow \: \: { \bold{ \int^b _{a} f(x).dx =\int^b _{a} f(a + b - x).dx }}\\$

• So that –

$\\ \implies \: \: { \bold{I= \int^2 _{-2} \left( {( - x)}^{3} \cos \dfrac{ - x}{2} + \dfrac{1}{2} \right) \sqrt{4 - {( - x)}^{2} }.dx }}\\$

$\\ \implies \: \: { \bold{I= \int^2 _{-2} \left( -{ x}^{3} \cos \dfrac{x}{2} + \dfrac{1}{2} \right) \sqrt{4 - {x}^{2} }.dx \:\:\:----eq.(2)}}\\$

• Now ADD eq.(1) & eq.(2) –

$\\ \implies \: \: { \bold{2I= \int^2 _{-2} \left \{ \left( -{ x}^{3} \cos \dfrac{x}{2} + \dfrac{1}{2} \right) \sqrt{4 - {x}^{2} } + \left( {x}^{3} \cos \dfrac{x}{2} + \dfrac{1}{2} \right) \sqrt{4 - {x}^{2} }\right\}.dx}}\\$

$\\ \implies \: \: { \bold{2I= \int^2 _{-2} \left( \dfrac{1}{2} + \dfrac{1}{2} \right) \sqrt{4 - {x}^{2} } .dx}}\\$

$\\ \implies \: \: { \bold{2I= \int^2 _{-2} \sqrt{4 - {x}^{2} } .dx}}\\$

• Using PROPERTY –

$\\ \dashrightarrow \: \: { \bold{\int^a _{-a} f(x).dx =2\int^a _{0} f(x).dx \: \: \: \: \: \:[ f(x) = even \: \: function]}}\\$

• So –

$\\ \implies \: \: { \bold{2I= 2\int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\$

$\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\$

• Now put x = 2 $\sin( \theta)$ –

$\\ \implies \: \: { \bold{x = 2 \sin( \theta)}} \\$

$\\ \implies \: \: { \bold{dx = 2 \cos( \theta).d \theta}} \\$

$\\ \implies \: \: { \bold{I= \int^ \frac{\pi}{2} _{0} \sqrt{4 - {(2 \sin \theta) }^{2} }(2 \cos\theta).d \theta}}\\$

$\\ \implies \: \: { \bold{I= \int^ \frac{\pi}{2} _{0} \sqrt{4 - {4 \sin^{2} \theta} }(2 \cos\theta).d \theta}}\\$

$\\ \implies \: \: { \bold{I= \int^ \frac{\pi}{2} _{0} 2 \sqrt{1 - {\sin^{2} \theta} }(2 \cos\theta).d \theta}}\\$

$\\ \implies \: \: { \bold{I= 2 \int^ \frac{\pi}{2} _{0} 2\cos^{2} \theta.d \theta}}\\$

$\\ \implies \: \: { \bold{I= 2 \int^ \frac{\pi}{2} _{0} 1 + \cos(2 \theta).d \theta}}\\$

$\\ \implies \: \: { \bold{I= 2 \left[ \theta+ \dfrac{ \sin(2 \theta)}{2} \right]^ \frac{\pi}{2} _{0} }}\\$

$\\ \implies \: \: { \bold{I= 2 \left[ \left( \dfrac{\pi}{2} - 0 \right)+ \dfrac{ \sin(2 \times \dfrac{\pi}{2})}{2} - \sin(0) \right]}}\\$

$\\ \implies \: \: { \bold{I= 2 \left[ \left( \dfrac{\pi}{2} \right)+ \dfrac{ \sin( \pi)}{2} - 0 \right]}}\\$

$\\ \implies \: \: { \bold{I= 2 \left[ \dfrac{\pi}{2} \right]}}\\$

$\\ \implies \large \: \:{ \boxed { \bold{I= \pi}}}\\$

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