$ \lim_{n \to \infty} \sum_{k=1}^n \sqrt{2- \left( \frac{k}{n} \right)^2} \cdot \frac{k}{n^2}$ is my solution any good?

Question

First write $$ \lim_{n \to \infty} \sum_{k=1}^n \sqrt{2- \left( \frac{k}{n} \right)^2} \cdot \frac{k}{n^2} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \sqrt{2- \left( \frac{k}{n} \right)^2} \cdot \frac{k}{n}.$$

From here I use the fact that $$ \lim_{n \to \infty} \frac{b-a}{n} \sum^n_{k=1}f(a + k \frac{b-a}{n})= \int^b_a f(x) \ dx, $$ to get $$ \lim_{n \to \infty} \sum_{k=1}^n \sqrt{2- \left( \frac{k}{n} \right)^2} \cdot \frac{k}{n^2} = \int^1_0 \sqrt{2-x^2} \ x \ dx. $$

Now write $u = 2 - x^2$, $du = -2x$, and $dx = -\frac{du}{2x}$

$u = 2- x^2$, so for $x = 0 \to u = 2 $ and for $ x = 1 \to u = 1$ $$ \int^1_0 \sqrt{2-x^2} \ x \ dx \to \int^2_1 \sqrt{u} \ x - \frac{du}{2x} = -\frac{1}{2}\int^2_1 \sqrt{u} \ du = -\frac{1}{2} \cdot \ \left[\frac{2}{3}u^{\frac{3}{2}} \right]^2_1 = \\ -\frac{1}{2} \cdot \ \left[\frac{2}{3}(2-x^2)^{\frac{3}{2}} \right]^1_0 = -\frac{1}{2} \cdot \ \left[\frac{2}{3}(2-1^2)^{\frac{3}{2}} \right] -\left(-\frac{1}{2} \cdot \ \left[\frac{2}{3}(2-0^2)^{\frac{3}{2}} \right] \right)= \frac{2\sqrt{2}-1}{3}. $$

Answer 1

For $I =\int^1_0 \sqrt{2-x^2} \ x \ dx $, if $u = 2-x^2$ then $du = -2xdx$ so $x dx = -du/2$ so $I = \int_{\sqrt{2}}^1 \sqrt{u}(-du/2) = \frac12\int_1^{\sqrt{2}} \sqrt{u}du =\frac12\dfrac{x^{3/2}}{3/2}|_1^{\sqrt{2}} =\frac13(2\sqrt{2}-1) $.