Math, asked by papafairy143, 6 hours ago

Solve the following integral

 \int_{0}^{\pi} (sin2x) dx

Answers

Answered by mathdude500
8

\large\underline{\sf{Solution-}}

Given integral is

\rm :\longmapsto\:\displaystyle\int_{0}^{\pi}\rm sin2x \: dx

Let assume that

\rm :\longmapsto\:I \:  =  \: \displaystyle\int_{0}^{\pi}\rm sin2x \: dx -  -  - (1)

We know,

 \purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{0}^{a}\rm f(x)dx = \displaystyle\int_{0}^{a}f(a - x) \: dx}}}

So, using this identity, we get

\rm :\longmapsto\:I \:  =  \: \displaystyle\int_{0}^{\pi}\rm sin2(\pi - x)\: dx

\rm :\longmapsto\:I \:  =  \: \displaystyle\int_{0}^{\pi}\rm sin(2\pi - 2x)\: dx

We know,

 \purple{\rm :\longmapsto\:\boxed{\tt{ sin(2\pi - x) = -  sinx \: }}}

So, using this, we get

\rm :\longmapsto\:I \:  =  \: -  \:  \displaystyle\int_{0}^{\pi}\rm sin2x \: dx

\rm :\longmapsto\:I \:  =  \: -  \:  I

\rm :\longmapsto\:2I \:  =  \: 0

\rm\implies \:I = 0

\rm\implies \:\:\displaystyle\int_{0}^{\pi}\rm sin2x \: dx \:  =  \: 0

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MORE TO KNOW

 \purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{a}^{b}\rm f(x)dx = \displaystyle\int_{a}^{b}f(y) \: dy}}}

 \purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{a}^{b}\rm f(x)dx =  \: -  \:  \displaystyle\int_{b}^{a}f(x) \: dx}}}

 \purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{a}^{b}\rm f(x)dx = \displaystyle\int_{a}^{b}f(a + b - x) \: dx}}}

 \purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{ - a}^{a}\rm f(x)dx = 0 \: if \: f( - x) =  - f(x)}}}

 \purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{ - a}^{a}\rm f(x)dx =2\displaystyle\int_{0}^{a}   \rm f(x) \: dx\: if \: f( - x) =  f(x)}}}

 \purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{0}^{2a}\rm f(x)dx = 0 \: if \: f(2a - x) =  - f(x)}}}

Answered by XxitzZBrainlyStarxX
10

Question:-

\sf \: Solve \:  the \:  following \:  integral \\  \sf \:  \int_{0}^{\pi} (sin2x) dx</p><p>

Solution:-

\sf \large \int_{0}^{\pi} (sin2x) dx

 \sf \large \longmapsto  \int_{0}^{\pi} (sin2x) dx

 \sf  \large \longmapsto  \bigg [  - \frac { 1}{2} cos2x \bigg]  ^\pi_0

 \sf \large \longmapsto  \bigg[ -  \frac{1}{2}  \: cos2\pi \bigg] -  \bigg[ -  \frac{1}{2}  \: cos \: 2 \times 0 \bigg]

 \sf \large \longmapsto \bigg[ -  \frac{1}{2}  \times 1 \bigg] -  \bigg[ -  \frac{1}{2}  \times 1 \bigg]

 \sf \large \longmapsto -  \frac{1}{2}  -  ( -  \frac{1}{2} )

 \sf \large \longmapsto{{ \cancel{  -  \frac{1}{2}}}}{{ \cancel{  +  \frac{1}{2} }}} = 0.

Hence, Answer is 0.

Answer:-

 \sf \large { \boxed{ \blue{\int_{0}^{\pi} (sin2x) dx=0}}}

Hope you have satisfied.

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