Here is the problem from Thomas Calculus book that I'm working on: Find $\displaystyle f\left(\frac{\pi}{2}\right)$ from the following information:
- $f$ is positive and continuous
- The area under the curve $y=f(x)$ from $x=0$ to $x=a$ is $$ \frac{a^2}{2}+\frac{a}{2}\sin a+\frac{\pi}{2}\cos a $$
My Work so far:
Given,
$$
\int_0^a f(t)dt=\frac{a^2}{2}+\frac{a}{2}\sin a+\frac{\pi}{2}\cos a \tag{2}
$$
This is where I say the problem statement is wrong, if $a=0$ then the above integral yields $\displaystyle \frac{\pi}{2}$, which shouldn't be the case because $$ \int_p^p f(t)dt=0 $$
I say, the value of the area should have been $$ \int_0^a f(t)dt=\frac{a^2}{2}+\frac{a}{2}\sin a+\frac{\pi}{2}\cos a-\frac{\pi}{2} $$
So, let $$ F(x)=\int_0^x f(t)dt=\frac{x^2}{2}+\frac{x}{2}\sin x+\frac{\pi}{2}\cos x-\frac{\pi}{2} $$
and $$ F'(x)=f(x)=x+\frac{1}{2}\left(\sin x+x\cos x\right)-\frac{\pi}{2}\sin x $$ and $$ f\left(\frac{\pi}{2}\right)=\frac{1}{2} $$
Integrating $f(x)$ between $x=0$ and $x=a$ and we end-up back at the modified area and I think I'm correct. Did I miss anything?
