Definite Integrals

Question

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Comments

ankitgupta.1729

@kumar.dilip

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@HeadShot , evaluation of $\int_{0}^{\frac{\pi}{2} }sin^{4}x\; cos^{5}x \;dx$ is correct but we can't get  2*$\int_{0}^{\frac{\pi}{2} }sin^{4}x\; cos^{5}x \;dx$ from $\int_{0}^{\pi}sin^{4}x\; cos^{5}x \;dx$ because here $f(2a-x) = f(x)$ is not true... That's why you are getting wrong answer here...

I have not seen such properties as $\int_{0}^{x}f(x) \; dx = k\int_{0}^{x/k} f(x)\;dx$ and $\int_{0}^{x} f(x)\;dx = \frac{1}{k} \int_{0}^{xk}\; f(x) \; dx$

if you have any source for these properties then kindly share...

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@ankitgupta.1729

 

Actually i dont have any standard reference though i ll verify and share with you.

Thanx for the help :)

Answer 1

Property of the definite integration$:$

$\int_{0}^{2a}f(x)dx=\left \{ 2\int_{0}^{a}f(x)dx ;f(2a-x)=f(x) \right \}$

$\int_{0}^{2a}f(x)dx=\left \{ 0;  f(2a-x)=-f(x) \right \}$

Now in your question $I=\int_{0}^{\pi }sin^4(x).cos^5(x)dx$

  $I=\int_{0}^{2(\frac{\pi}{2}) }sin^4(x).cos^5(x)dx$

Here $f(x)=sin^4(x).cos^5(x)$

Now,find$f(\pi-x)=sin^4(\pi-x).cos^5(\pi-x)$       

             $f(\pi-x)=sin^4(x.)(-cos^5(x))$        $[cos(\pi-x) = -cos(x) ,sin(\pi-x) = sin(x)]$

             $f(\pi-x)=-sin^4(x.)cos^5(x)=-f(x)$ 

So,$I=\int_{0}^{\pi }sin^4(x).cos^5(x)dx=0$  [From the above property]