Question

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Answer 1

GIVEN :–

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

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 2

Answer:

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