Question

please answer my question... ​

Answer 1

Answer:

but the ki Bhagalpur Bhandara Bharatpur Bharuch Gujarat India society of

Answer 2

Answer:

⟹2I=2∫

⟹2I=2∫ 0

⟹2I=2∫ 02

⟹2I=2∫ 02

⟹2I=2∫ 02

⟹2I=2∫ 02 4−x

⟹2I=2∫ 02 4−x 2

⟹2I=2∫ 02 4−x 2

⟹2I=2∫ 02 4−x 2

⟹2I=2∫ 02 4−x 2 .dx

⟹2I=2∫ 02 4−x 2 .dx

⟹2I=2∫ 02 4−x 2 .dx

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered}

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫ 0

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫ 02

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫ 02

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫ 02

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫ 02 4−x

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫ 02 4−x 2

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫ 02 4−x 2

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫ 02 4−x 2

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫ 02 4−x 2 .dx

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫ 02 4−x 2 .dx

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫ 02 4−x 2 .dx

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫ 02 4−x 2 .dx • Now put x = 2 sin⁡(θ)\sin( \theta)sin(θ) –

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫ 02 4−x 2 .dx • Now put x = 2 sin⁡(θ)\sin( \theta)sin(θ) –

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫ 02 4−x 2 .dx • Now put x = 2 sin⁡(θ)\sin( \theta)sin(θ) –

⟹2I=2∫ 02 4−x 2 .dx ⟹I=∫024−x2.dx\begin{lgathered}\\ \implies \: \: { \bold{I= \int^2 _{0} \sqrt{4 - {x}^{2} } .dx}}\\\end{lgathered} ⟹I=∫ 02 4−x 2 .dx • Now put x = 2 sin⁡(θ)\sin( \theta)sin(θ) –