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$$ \int_2^4\frac{\sqrt{\log(9-x)}}{\sqrt{\log(9-x)}+\sqrt{\log(3+x)}}dx=1$$

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In general: $\int_2^4 \frac{f(x)}{f(x)+f(6-x)}\,dx=1$. –  pharmine Nov 10 '11 at 5:22
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why is this the case? –  vivaelche05 Nov 10 '11 at 5:26
    
I posted an answer below. –  pharmine Nov 10 '11 at 5:50

2 Answers 2

up vote 5 down vote accepted

We show that $I=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}\,dx$ equals $1$, where $f(x)=\sqrt{\log(9-x)}$.

Proof. By making a substitution $y=6-x$, we get $$I=\int_4^2 -\frac{f(6-y)}{f(y)+f(6-y)}\,dy=\int_2^4 \frac{f(6-y)}{f(y)+f(6-y)}\,dy.$$

Therefore $$\begin{align*}2I&=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}dx+\int_2^4 \frac{f(6-y)}{f(y)+f(6-y)}dy\\ &=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}dx+\int_2^4 \frac{f(6-x)}{f(x)+f(6-x)}dx=\int_2^4 1\,dx =2.\end{align*}$$

Edit. In the last part, $y$ is a dummy variable and can be changed to $x$ or any other variable you like.

Edit2. If you don't like the same $x$ being used, you could use another variable, say $s$, so that with two substitutions $$\begin{align*}2I&=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}dx+\int_2^4 \frac{f(6-y)}{f(y)+f(6-y)}dy\\ &=\int_2^4 \frac{f(s)}{f(s)+f(6-s)}ds+\int_2^4 \frac{f(6-s)}{f(s)+f(6-s)}ds=\int_2^4 1\,ds =2.\end{align*}$$

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how did f(6-y) become f(6-x) in the numerator of the second integrand on the last line? –  vivaelche05 Nov 10 '11 at 5:50
    
Because $y$ is a dummy variable; please see my edited comment. –  pharmine Nov 10 '11 at 5:53
    
Wow, thank you. can I ask, conceptually, why this is true for all integrable f(x)'s? –  vivaelche05 Nov 10 '11 at 15:18
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As Dinesh points out, $g(x)=\frac{f(x)}{f(x)+f(6-x)}$ satisfies $g(x)+g(6-x)=1$ (more generally, $g(x)+g(a-x)=b$ where $a$ and $b$ are constants); this is the condition where you can use this integration trick. –  pharmine Nov 10 '11 at 15:33

$ \int^a_b f(x)\,dx $ = $ \int^a_b f(a+b-x)\,dx $ We can prove this by changing the dummy variable x to a+b-x, we get the integrand as $-f(a+b-x)\,dx$ and by changing the limits accordingly with $a+b-b$ and $a+b-a$, i.e. Now the integral becomes $-\int^b_af(a+b-x)dx$=$ \int^a_b f(a+b-x)dx $

The function given in the question satisfies the property that $f(x)+f(6-x)=1$. Let the integral be $I$ then $2I= \int^a_b f(x)\,dx+ \int^a_b f(a+b-x)dx=\int^a_b f(x)+f(a+b-x)\,dx=\int^a_b\,dx=a-b$, In your case $a-b=2$, hence $2I=2$. So $I=1$.

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