You can calcuate these two sums independently:
$$\int_{-5}^{5} (x-\sqrt{25-x^2})dx=\int_{-5}^{5} xdx-\int_{-5}^{5} \sqrt{25-x^2}dx$$
Let's calculate the first integral. Obviously it's zero because the function $f(x)=x$ is odd but if you insist you can prove it by summation:
Divide interval from -5 to +5 in $n$ equal segments:
$$\Delta x_i=\frac{10}n, \quad y_i=x_i=-5+\frac {10i}n,\quad i=0, 1, 2,...,n $$
$$A_n=\sum_{i=0}^{n-1}y_i\Delta x_i=\sum_{i=0}^{n-1}(-5+\frac {10i}n)\frac{10}n$$
$$=-50\sum_{i=0}^{n-1}\frac1n+\frac{100}{n^2}\sum_{i=0}^{n-1}i=-50+\frac{100}{n^2}\frac{n(n-1)}{2}=-50+50\frac{n-1}n=-\frac{50}n$$
So the first integral is:
$$I_1=\lim_{n\to \infty}A_n=\lim_{n\to \infty}\frac{-50}{n}=0$$
Let's tackle the second one:
$$I_2=\int_{-5}^{5} \sqrt{25-x^2}dx$$
The graph of function $f(x)=\sqrt{25-x^2}$ is symmetric with respect to $y$-axis so:
$$I_2=2\int_{0}^{5} \sqrt{25-x^2}dx$$
Let's calculate:
$$I_3=\int_{0}^{5} \sqrt{25-x^2}dx$$
...by symmation. Again, I will divide the interval of integration into $n$ segments but this time they won't be of equal length. I will introduce variable $\varphi$ such that:
$$\Delta\varphi=\frac{\pi}{2n}, \quad \varphi_i=i\Delta\varphi=\frac{i\pi}{2n}, \quad i=0,1,2,...,n$$
$$x_i=5\sin\varphi_i$$
$$y_i=\sqrt{25-x_i^2}=5\cos\varphi_i$$
$$\Delta x_i=x_{i+1}-x_i=5(\sin\varphi_{i+1}-\sin\varphi_{i})=5(\sin(\varphi_i+\Delta\varphi)-\sin\varphi_i)=5(\sin\varphi_i(\cos\Delta\varphi-1)+\cos\varphi_i\sin\Delta\varphi)=5(-2\sin^2\frac{\Delta\varphi}{2}\sin\varphi_i+\cos\varphi_i\sin\Delta\varphi)$$
Having in mind that for small values of $\Delta\varphi$:
$$\sin \Delta\varphi=\Delta\varphi+O(\Delta\varphi^3)$$
...we can write:
$$\Delta x_i=5\cos\varphi_i\Delta\varphi$$
Now the integral becomes:
$$\Delta A_i=y_i\Delta x_i=25\Delta\varphi\cos^2\varphi_i=\frac{25\Delta\varphi}{2}(1+cos{2\varphi_i})=\frac{25\Delta\varphi}{2}(1+\cos\frac{i\pi}{n})$$
$$A_n=\sum_{i=0}^n \Delta A_i=\frac{25\Delta\varphi}{2}\sum_{i=0}^{n-1}1+\frac{25\Delta\varphi}{2}\sum_{i=0}^{n-1} \cos\frac{i\pi}{n}=\frac{25n\Delta\varphi}{2}+\frac{25\Delta\varphi}{2}\sum_{i=0}^{n-1} \cos\frac{i\pi}{n}$$
$$=\frac{25\pi}{4}+\frac{25\pi}{4n}\sum_{i=0}^{n-1} \cos\frac{i\pi}{n}$$
By using trigonometric identity of Lagrange you can show that:
$$\sum_{i=0}^{n-1} \cos\frac{i\pi}{n}=1+\sum_{i=1}^{n-1} \cos\frac{i\pi}{n}=1-\frac12+\frac{\sin((n-1+\frac12)\frac{\pi}{n})}{2\sin\frac{\pi}{2n}}=\frac12+\frac{\sin(\pi-\frac{\pi}{2n})}{2\sin\frac{\pi}{2n}}=1$$
It means that:
$$A_n=\frac{25\pi}{4}+\frac{25\pi}{4n}$$
$$I_3=\lim_{n\to\infty}A_n=\lim_{n\to\infty}\frac{25\pi}{4}+\frac{25\pi}{4n}=\frac{25\pi}{4}$$
$$I_2=2I_3=\frac{25\pi}{2}$$
Best Answer
HINT:
Using this, $$\sum_{1\le r\le n}\sin \frac{\pi r}n=\frac{\sin \left( n\cdot \frac{\pi}{2n}\right)}{\sin \frac \pi{2n}}\cdot \sin \left(\frac{2\frac\pi n+(n-1)\frac{\pi}n}2\right)=\frac{\sin\left(\frac{(n+1)\pi}{2n}\right)}{\sin \frac \pi{2n}}$$
Now, $\sin\left(\frac{(n+1)\pi}{2n}\right)=\sin\left(\frac\pi2+\frac\pi{2n}\right)=\cos \frac\pi{2n}\implies \lim_{n\to\infty}\sin\left(\frac{(n+1)\pi}{2n}\right)=1$
$$\lim_{n\to\infty}\frac 1{n\sin \frac \pi{2n}}=\lim_{h\to0}\frac {2h}{\pi\sin h }(\text{ putting } \frac \pi{2n}=h)$$
$$\implies \lim_{n\to\infty}\frac 1{n\sin \frac \pi{2n}}=\frac2\pi\cdot\lim_{h\to0}\frac h{\sin h }=\frac2\pi$$