Evaluate the following integral where it is noted that the denominator never vanishes over the interval. No assumption is made on the continuity of f(x) or its derivatives.
$$\int_{0}^{a} \frac{f(x)}{f(x)+f(a-x)} dx$$
Asked
Active
Viewed 81 times
0
-
2This uses a classic trick: substitute $u=a-x$. Then look carefully at what you've got. I won't reveal more than that... ;) – Feb 18 '18 at 00:05
-
@user8734617 I think I spoiled the final. – Enrico M. Feb 18 '18 at 00:06
-
1At least they use MathJax... – Yuriy S Feb 18 '18 at 00:06
-
1@VonNeumann Not at all. There is still some work for the OP to do. – Feb 18 '18 at 00:07
-
1Welcome to stackexchange. You are more likely to get useful answers rather than downvotes and votes to close if you edit the question to show what you tried and where you are stuck. – Ethan Bolker Feb 18 '18 at 00:11
-
A somewhat easier solution is given below. – donal obaoill Feb 18 '18 at 00:45
-
Some may consider that the classic answer is a non obvious use of substitution of variables, however by going back to a more fundamental approach the solution soon reveals itself. – donal obaoill Feb 18 '18 at 00:48
2 Answers
2
$$\int_{0}^{a} \frac{f(x)}{f(x)+f(a-x)} dx$$
Consider that the interval of integration is divided up into n equal segments with n+1 end points $x_0,x_1,..x_n.$ The y values are given by $y_0=f(x_0),y_1=f(x_1)...y_n=f(x_n)=f(a).$
$$\int_{0}^{a} \frac{f(x)}{f(x)+f(a-x)} dx$$ can be approximated by the sum $$\frac{a}{n} \left( \frac{y_0}{y_0+y_n}+\frac{y_1}{y_1+y_{n-1}}+ \ldots \frac{y_n}{y_0+y_n} \right)$$ Thus there are n/2 fractions that have the same denominator which can be combined to obtain $$\frac{a}{n} \left( \frac{y_0+y_n}{y_0+y_n}+\frac{y_1+y_{n-1}}{y_1+y_{n-1}} + \ldots \right) =\frac{a}{n} \frac{n}{2}=\frac{a}{2} $$ Notice that the n disappears from the summation. Even if the number of terms was infinite the answer is still the same. This means that not only is the integral equal to this value but also the discrete sum.