Animated Solution for Mathematics - Definite Integration: If ${\int }_0^2(\sqrt{2x}-\sqrt{2x-x^2})dx={\int }_0^1(1-\sqrt{1-y^2}-\frac{y^2}{2})dy+{\int }_1^2(2-\frac{y^2}{2})dy+I$, then $I$ is equal to

Welcome, fellow traveler of the JEE path. Today, we aren't just solving an integral; we are peeling back the layers of a geometric puzzle.

When you look at an expression like ${\int }_0^2(\sqrt{2x}-\sqrt{2x-x^2})dx$, it is easy to feel overwhelmed. But I want you to pause and look past the symbols. You are observing the difference between two areas: one bounded by a parabola and the other—that mysterious $\sqrt{2x-x^2}$—representing the heartbeat of a circle.

The Parabola and the Circle

Let us tackle the Left Hand Side (LHS) first. We split this into two parts: $L_1={\int }_0^2\sqrt{2x}dx$ and $L_2={\int }_0^2\sqrt{2x-x^2}dx$.

For $L_1$, we use the power rule:

Now, for $L_2$. If you complete the square, $2x-x^2$ becomes $1-(x-1{)}^2$.

The integral ${\int }_0^2\sqrt{1-(x-1{)}^2}dx$ reveals itself as the area of a semi-circle with radius $r=1$ centered at $(1,0)$. Since the area of a full circle is $\pi r^2$, our semi-circle is simply $\frac{\pi }{2}$. Thus, our total LHS is $\frac{8}{3}-\frac{\pi }{2}$.

The RHS Decomposition

Now, look at the Right Hand Side. It is split into two integrals, $R_1$ and $R_2$, plus our unknown $I$.

We evaluate $R_1={\int }_0^1(1-\sqrt{1-y^2}-\frac{y^2}{2})dy$. By splitting this into three distinct terms, we find:

Next, $R_2={\int }_1^2(2-\frac{y^2}{2})dy$. This is a standard polynomial integral:

The Final Synthesis

We have our LHS: $\frac{8}{3}-\frac{\pi }{2}$. We have our RHS: $(\frac{5}{6}-\frac{\pi }{4})+\frac{5}{6}+I$.

Combining the constants on the RHS, we get $\frac{5}{3}-\frac{\pi }{4}+I$. Now, we set them equal:

Isolating $I$ is now a matter of careful arithmetic. Subtracting $\frac{5}{3}$ from $\frac{8}{3}$ gives $1$, and adding $\frac{\pi }{4}$ to $-\frac{\pi }{2}$ leaves us with $-\frac{\pi }{4}$. Therefore, $I=1-\frac{\pi }{4}$.

Finally, we compare this to our options. When we evaluate ${\int }_0^1(1-\sqrt{1-y^2})dy$, we get $[y{]}_0^1-{\int }_0^1\sqrt{1-y^2}dy$, which is exactly $1-\frac{\pi }{4}$.

See? The complexity was just a mask. By breaking the problem into geometric components and handling the arithmetic with patience, we didn't just find the answer—we understood the landscape of the function.