Differentiate $y = \frac{{x\sqrt {{a^2} - {x^2}} }}{2} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\frac{x}{a}$ with respect to x, Integrate the result so obtained. Compare the two results and draw conclusion from your results.

To differentiate the given function, we can use the product rule, the chain rule, and the derivative of the inverse sine function. The product rule states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$. The chain rule states that if $f(x) = g(h(x))$, then $f'(x) = g'(h(x))h'(x)$. The derivative of the inverse sine function is $\frac{d}{dx} \sin^{-1} x = \frac{1}{\sqrt{1-x^2}}$. Using these rules, we can find the derivative of $y$ as follows: $$ \begin{aligned} y &= \frac{{x\sqrt {{a^2} - {x^2}} }}{2} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\frac{x}{a} \\ y' &= \frac{1}{2} \left( \sqrt{a^2 - x^2} + x \frac{d}{dx} \sqrt{a^2 - x^2} \right) + \frac{a^2}{2} \frac{d}{dx} \sin^{-1} \frac{x}{a} \\ &= \frac{1}{2} \left( \sqrt{a^2 - x^2} + x \frac{1}{2} (a^2 - x^2)^{-1/2} (-2x) \right) + \frac{a^2}{2} \frac{1}{\sqrt{1 - (x/a)^2}} \frac{1}{a} \\ &= \frac{1}{2} \left( \sqrt{a^2 - x^2} - \frac{x^2}{\sqrt{a^2 - x^2}} \right) + \frac{1}{2} \frac{1}{\sqrt{a^2 - x^2}} \\ &= \frac{1}{2} \frac{a^2 - 2x^2}{\sqrt{a^2 - x^2}} + \frac{1}{2} \frac{1}{\sqrt{a^2 - x^2}} \\ &= \frac{1}{2} \frac{a^2 - x^2}{\sqrt{a^2 - x^2}} \end{aligned} $$ To integrate the result, we can use the substitution method. Let $u = a^2 - x^2$, then $du = -2x dx$. The integral becomes: $$ \begin{aligned} \int y' dx &= \int \frac{1}{2} \frac{u}{\sqrt{u}} \frac{-du}{2x} \\ &= -\frac{1}{4} \int \frac{u}{\sqrt{u}} \frac{du}{x} \\ &= -\frac{1}{4} \int \sqrt{u} du \\ &= -\frac{1}{4} \frac{2}{3} u^{3/2} + C \\ &= -\frac{1}{6} (a^2 - x^2)^{3/2} + C \end{aligned} $$ To compare the two results, we can use the fundamental theorem of calculus, which states that if $F(x)$ is an antiderivative of $f(x)$, then $\int_a^b f(x) dx = F(b) - F(a)$. In our case, $F(x) = -\frac{1}{6} (a^2 - x^2)^{3/2} + C$ is an antiderivative of $f(x) = y' = \frac{1}{2} \frac{a^2 - x^2}{\sqrt{a^2 - x^2}}$. Therefore, we have: $$ \int_a^b y' dx = F(b) - F(a) = -\frac{1}{6} (a^2 - b^2)^{3/2} + C - \left( -\frac{1}{6} (a^2 - a^2)^{3/2} + C \right) = -\frac{1}{6} (a^2 - b^2)^{3/2} + \frac{1}{6} (a^2)^{3/2} $$ On the other hand, we can evaluate $y$ at the endpoints $a$ and $b$ as follows: $$ \begin{aligned} y(a) &= \frac{{a\sqrt {{a^2} - {a^2}} }}{2} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\frac{a}{a} \\ &= 0 + \frac{a^2}{2} \sin^{-1} 1 \\ &= \frac{a^2}{2} \frac{\pi}{2} \\ &= \frac{a^2 \pi}{4} \\ y(b) &= \frac{{b\sqrt {{a^2} - {b^2}} }}{2} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\frac{b}{a} \\ &= \frac{b}{2} \sqrt{a^2 - b^2} + \frac{a^2}{2} \sin^{-1} \frac{b}{a} \end{aligned} $$ Therefore, we have: $$ y(b) - y(a) = \frac{b}{2} \sqrt{a^2 - b^2} + \frac{a^2}{2} \sin^{-1} \frac{b}{a} - \frac{a^2 \pi}{4} $$ We can see that the two expressions are not equal in general, but they are equal when $b = 0$ or $b = a$. This means that the function $y$ satisfies the differential equation $y' = \frac{1}{2} \frac{a^2 - x^2}{\sqrt{a^2 - x^2}}$ only when $x = 0$ or $x = a$. Otherwise, there is a discrepancy between the derivative and the integral of $y$. This discrepancy can be explained by the fact that the function $y$ is not continuous at $x = a$, since it has a vertical asymptote there. Therefore, the fundamental theorem of calculus does not apply to $y$ on the interval $[0, a]$, and we cannot use the antiderivative of $y'$ to find the value of $y$. A possible conclusion from this result is that we need to be careful when applying the fundamental theorem of calculus to functions that are not continuous or differentiable on the interval of integration.