JEE Main 2022 — Mathematics Calculus

Plotting the diagram of given data we have,

Now finding limit by diagram, we get

$${x}^{\frac{2}{3}}+{x}^{\frac{2}{3}}=1$$x32​+x32​=1

$$\Rightarrow x=\pm \frac{1}{2\sqrt{2}}$$⇒x=±22​1​

So total area will be given by,

$$A={\int }_{-\frac{1}{2\sqrt{2}}}^{0}{(1-{x}^{\frac{2}{3}})}^{\frac{3}{2}}dx-(\frac{1}{2}\times \frac{1}{2\sqrt{2}}\times \frac{1}{2\sqrt{2}})+{\int }_{0}^{1}{(1-{x}^{\frac{2}{3}})}^{\frac{3}{2}}dx$$A=∫−22​1​0​(1−x32​)23​dx−(21​×22​1​×22​1​)+∫01​(1−x32​)23​dx

Now for solving $\displaystyle \int {(1-{x}^{\frac{2}{3}})}^{\frac{3}{2}}dx$∫(1−x32​)23​dx

Let $x={\mathrm{sin}}^{3}\theta$x=sin3θ

$$=\int {(1-{\mathrm{sin}}^{2}\theta )}^{\frac{3}{2}}\cdot 3{\mathrm{sin}}^{2}\theta \mathrm{cos}\theta d\theta$$=∫(1−sin2θ)23​⋅3sin2θcosθdθ

$$=\int 3{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{4}\theta d\theta$$=∫3sin2θcos4θdθ

$$=3\int {\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{4}\theta d\theta$$=3∫sin2θcos4θdθ

Now putting the limit and solving we get

$$\Rightarrow A=\frac{9\pi }{64}-\frac{1}{16}+\frac{1}{16}=\frac{36\pi }{256}$$⇒A=649π​−161​+161​=25636π​

$$\Rightarrow \frac{256A}{\pi }=36$$⇒π256A​=36