In the subject test of 1987 problem 30 asks:
- The improper integral $\displaystyle \int_a^b f(x) f'(x) \, dx$ is
A.) necessarily zero
B.) possibly zero but not necessarily
C.) necessarily nonexistent
D.) possibly nonexistent but not necessarily
E.) none of the above
The accepted answer was (A.)
My question, why is $\int_0^1 f(x) f'(x) dx = 1/2$ for $ f(x) = \sqrt{x}$ not a counter-example? I mean, $1/2 \neq 0$, this much is clear.
How should we think about this question? Is my integral not genuinely improper so it doesn't count?
Thanks in advance for your help.
Did you note that there is an instruction above problem 28 that gives information applying to problems 28 - 30?
The function is a semicircle with endpoints at $(a,0)$ and $(b,0)$ with $a<b$.
http://www.math.ucla.edu/~cmarshak/GRE3.pdf
We want to show that $\int_a^b f(x)f'(x)dx=0$, with $f$ the semicircle described above. This means $f(x)=\sqrt{(\frac{b-a}{2})^2-(x-\frac{a+b}{2})^2}$.
Then $f'(x)=\frac{-2(x-\frac{a+b}{2})}{2f(x)}$.
So $f(x)f'(x)=-x+\frac{a+b}{2}$.
Then $\int_a^b f(x)f'(x) dx=\int_a^b(-x+\frac{a+b}{2})dx=[-\frac{x^2}{2}+\frac{(a+b)x}{2}]_a^b=-\frac{b^2-a^2}{2}+\frac{(a+b)(b-a)}{2}=0$