JEE Main 2022 — Definite Integration Question with Solution

Given,

$${\int }_0^2\left(\sqrt{2x}-\sqrt{2x-x^2}\right)dx={\int }_0^1\left(1-\sqrt{1-y^2}-\frac{y^2}{2}\right)dy+{\int }_1^2\left(2-\frac{y^2}{2}\right)dy+I$$

Now,

$$L.H.S.={\int }_0^2\left(\sqrt{2x}-\sqrt{2x-x^2}\right)dx$$

$$=\sqrt{2}{\int }_0^2{x}^{1/2}dx-{\int }_0^2\sqrt{-(x^2-2x)}dx$$

$$=\sqrt{2}{\left[\frac{x^{3/2}}{\frac{3}{2}}\right]}_0^2-{\int }_0^2\sqrt{-\left[{(x-1)}^2-1\right]}dx$$

$$=\sqrt{2}\times \frac{2}{3}\left[2^{\frac{3}{2}}\right]-{\int }_0^2\sqrt{1-{(x-1)}^2}dx$$

[We know,

$$\sqrt{a^2-x^2}=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)+C$$]

$$=\frac{2\sqrt{2}\times 2\sqrt{2}}{3}-{\left[\frac{x-1}{2}\sqrt{1-{(x-1)}^2}+\frac{1}{2}{\sin }^{-1}\left(\frac{x-1}{1}\right)\right]}_0^2$$

$$=\frac{8}{3}-\left[\frac{1}{2}\sqrt{1-1}+\frac{1}{2}{\sin }^{-1}(1)-\left(\left(\frac{1}{2}\right)\sqrt{1-1}+\frac{1}{2}{\sin }^{-1}(-1)\right)\right]$$

$$=\frac{8}{3}-\frac{1}{2}{\sin }^{-1}\left(\sin \frac{\pi }{2}\right)+\frac{1}{2}{\sin }^{-1}\left(\sin \frac{3\pi }{2}\right)$$

$$=\frac{8}{3}-\frac{1}{2}.\frac{\pi }{2}+\frac{1}{2}.\frac{3\pi }{2}$$

$$=\frac{8}{3}-\frac{\pi }{4}+\frac{3\pi }{4}$$

$$=\frac{8}{3}-\frac{2\pi }{4}$$

$$=\frac{8}{3}-\frac{\pi }{2}$$

Now in R.H.S.,

$${\int }_0^1\left(1-\sqrt{1-y^2}-\frac{y^2}{2}\right)dy$$

$$={\int }_0^{1}dy-{\int }_0^1\sqrt{1-y^2}-{\int }_0^1\frac{y^2}{2}dy$$

$$={\left[y\right]}_0^1-{\left[\frac{y}{2}\sqrt{1-y^2}+\frac{1}{2}{\sin }^{-1}\left(\frac{y}{1}\right)\right]}_0^1-{\left[\frac{y^3}{6}\right]}_0^1$$

$$=1-\left[\left(\frac{1}{2}\sqrt{1-0}+\frac{1}{2}{\sin }^{-1}(1)\right)-\left(0+{\sin }^{-1}(0)\right)\right]-\frac{1}{6}$$

$$=1-\frac{1}{2}{\sin }^{-1}\left(\sin \frac{\pi }{2}\right)-\frac{1}{6}$$

$$=1-\frac{1}{2}\times \frac{\pi }{2}-\frac{1}{6}$$

$$=\frac{5}{6}-\frac{\pi }{4}$$

Now,

$${\int }_1^2\left(2-\frac{y^2}{2}\right)dy$$

$$={\left[2y-\frac{y^3}{6}\right]}_1^2$$

$$=\left[\left(4-\frac{8}{6}\right)-\left(2-\frac{1}{6}\right)\right]$$

$$=4-\frac{8}{6}-2+\frac{1}{6}$$

$$=2-\frac{7}{6}$$

$$=\frac{5}{6}$$

$$\therefore$$ $$R.H.S.=\frac{5}{6}-\frac{\pi }{4}+\frac{5}{6}+I$$

As $$L.H.S.=R.H.S.$$

$$\Rightarrow \frac{8}{3}-\frac{\pi }{2}=\frac{10}{6}-\frac{\pi }{4}+I$$

$$\Rightarrow \frac{10}{6}-\frac{8}{3}+I=-\frac{\pi }{2}+\frac{\pi }{4}$$

$$\Rightarrow \frac{-6}{6}+I=-\frac{\pi }{4}$$

$$\Rightarrow I=1-\frac{\pi }{4}$$

From option (c)

$${\int }_0^1(1-\sqrt{1-y^2})dy$$

$$={\left[y\right]}_0^1-{\int }_0^1(\sqrt{1-y^2})dy$$

$$=1-{\left[\frac{y}{2}\sqrt{1-y^2}+\frac{1}{2}{\sin }^{-1}\left(\frac{y}{1}\right)\right]}_0^1$$

$$=1-\left[\frac{1}{2}\sqrt{1-1}+\frac{1}{2}{\sin }^{-1}(1)-\left(0+\frac{1}{2}{\sin }^{-1}(0)\right)\right]$$

$$=1-\left[\frac{1}{2}\times \frac{\pi }{2}-0\right]$$

$$=1-\frac{\pi }{4}=I$$

$$\therefore$$ Option (c) is the right answer.

Note :

There is no way to guess which option is correct. You have to check all the options to see which give value equal to I.