Integrate The Product

Calculus Level pending

Each of q ( y ) q(y) , r ( y ) r(y) and s ( y ) s(y) are continuous functions on [ 0 , 6 ] [0, 6] such that:

q ( y ) = q ( 6 − y ) q(y) = q(6-y) , r ( y ) = − r ( 6 − y ) r(y) = - r(6-y) , and 3 s ( y ) − 4 s ( 6 − y ) = 5 3s(y) - 4s(6-y) = 5

Evaluate: ∫ 0 6 q ( y ) r ( y ) s ( y ) d y \displaystyle\int_{0}^{6}q(y)r(y)s(y)dy


The answer is 0.

Danish Ahmed by Danish Ahmed , 24, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Rishabh Jain
Jan 26, 2016

Using ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x \displaystyle\int_{a}^{b}f(x)dx=\displaystyle\int_{a}^{b} f(a+b-x)dx H e n c e I = ∫ 0 6 q ( y ) r ( y ) s ( y ) d y = ∫ 0 6 q ( 6 − y ) r ( 6 − y ) s ( 6 − y ) d y = ∫ 0 6 q ( y ) ( − r ( y ) ) ( 3 s ( y ) − 5 4 ) d y = − 3 I + 5 ( ∫ 0 6 q ( y ) r ( y ) d y ) 4 ⇒ I = − 3 I 4 + 0 See Note ⇒ I = 0 Hence~I=\displaystyle\int_{0}^{6}q(y)r(y)s(y)dy \\=\displaystyle\int_{0}^{6}q(6-y)r(6-y)s(6-y)dy\\ =\displaystyle\int_{0}^{6}q(y)(-r(y))(\frac{3s(y)-5}{4})dy\\ =\dfrac{-3I + 5(\displaystyle\int_{0}^{6}q(y)r(y)dy)}{4}\\ \Rightarrow I=\dfrac{-3I}{4}+0~~~~\color{#D61F06}{\text{See Note}}\\ \Rightarrow I=0 Using same property ∫ 0 6 q ( y ) r ( y ) d y = − ∫ 0 6 q ( y ) r ( y ) d y ⇒ ∫ 0 6 q ( y ) r ( y ) d y = 0 \small{\color{#D61F06}{\text{Using same property}\displaystyle\int_{0}^{6}q(y)r(y)dy =-\displaystyle\int_{0}^{6}q(y)r(y)dy\\ \Rightarrow \displaystyle\int_{0}^{6}q(y)r(y)dy=0}}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...