One statement about definite integrals implying an another.

Question

Could anyone please explain me the following statements :

"$$\int_{0}^{\theta}g(y)\frac{2n}{\theta^{2n}}y^{2n-1} dy=0$$ for all $\theta$ implies $$g(\theta)\frac{2n}{\theta^{2n}}\theta^{2n-1}=0$$ for all $\theta$ by taking derivatives. This can only be zero if $g(\theta)=0$ for all $\theta$."

Answer 1

That's the fundamental theorem of calculus. If

$$f(\theta):=\int_{0}^{\theta}g(y)\frac{2n}{\theta^{2n}}y^{2n-1} dy=0$$

for all $\theta$, then we differentiate to get

$$\begin{split} 0 &= f'(\theta)\\ & =g(\theta)\frac{2n}{\theta^{2n}}\theta^{2n-1} - 2n \int_0^\theta g(y)\frac{2n}{\theta^{2n+1}}y^{2n-1} dy\\ &= g(\theta)\frac{2n}{\theta^{2n}}\theta^{2n-1} - \frac{2n}{\theta} \int_{0}^{\theta}g(y)\frac{2n}{\theta^{2n}}y^{2n-1} dy \\ &= g(\theta)\frac{2n}{\theta^{2n}}\theta^{2n-1}. \end{split}$$