JEE Main 2022 — Definite Integration Question with Solution

$${\int }_0^1\left[2x-|3x^2-5x+2|+1\right]dx$$

$$3x^2-5x+2=0$$

$$\Rightarrow 3x^2-3x-2x+2=0$$

$$\Rightarrow 3x(x-1)-2(x-1)=0$$

$$\Rightarrow (x-1)(3x-2)=0$$

$$\therefore$$ In 0 to $$\frac{2}{3},|3x^2-5x+2|=3x^2-5x+2$$

And in $$\frac{2}{3}$$ to 1, $$|3x^2-5x+2|=-(3x^2-5x+2)$$

$$\therefore$$ $${\int }_0^{\frac{2}{3}}\left[2x-3x^2+5x-2+1\right]dx+{\int }_{\frac{2}{3}}^1\left[2x+3x^2-5x+2+1\right]dx$$

$$={\int }_0^{\frac{2}{3}}\left[-3x^2+7x-1\right]dx+{\int }_{\frac{2}{3}}^1\left[3x^2-3x+3\right]dx$$

Now, let $$f(x)=-3x^2+7x-1$$

$$\Rightarrow f^{\prime }(x)=-6x+7$$

$$=-6\left(x-\frac{7}{6}\right)$$

$$f(0)=-1$$

$$f\left(\frac{2}{3}\right)=-3\times \frac{4}{9}+\frac{14}{3}-1$$

$$=\frac{7}{3}=2.33$$

$$\therefore$$ Range of $$f(x)=\left[-1,\frac{7}{3}\right]$$

Integer value between $$-1$$ to $$\frac{7}{3}=-1,0,1,2,\frac{7}{3}$$

When $$f(x)=-1$$

$$\Rightarrow -3x^2+7x-1=-1$$

$$\Rightarrow 3x^2-7x=0$$

$$\Rightarrow x(3x-7)=0$$

$$\Rightarrow x=0,\frac{7}{3}$$ (not possible as $$\frac{7}{3}\notin (0,2)$$)

When $$f(x)=0$$

$$\Rightarrow -3x^2+7x-1=0$$

$$\Rightarrow 3x^2-7x+1=0$$

$$\Rightarrow x=\frac{7\pm \sqrt{49-12}}{6}$$

$$=\frac{7\pm \sqrt{37}}{6}$$

$$\therefore$$ $$x=\frac{7-\sqrt{37}}{6}$$

When $$f(x)=1$$

$$\Rightarrow -3x^2+7x-1=1$$

$$\Rightarrow 3x^2-7x+2=0$$

$$\Rightarrow x=\frac{7\pm \sqrt{49-24}}{6}$$

$$=\frac{7\pm \sqrt{25}}{6}$$

$$=\frac{7\pm 5}{6}$$

$$\Rightarrow x=2,\frac{1}{3}$$

When $$f(x)=2$$

$$\Rightarrow -3x^2+7x-1=2$$

$$\Rightarrow 3x^2-7x+3=0$$

$$\Rightarrow x=\frac{7\pm \sqrt{49-36}}{6}$$

$$=\frac{7\pm \sqrt{13}}{6}$$

$$\therefore$$ $$x=\frac{7-\sqrt{13}}{6}$$

$$\therefore$$ Possible values of $$x=0,\frac{7-\sqrt{37}}{6},\frac{1}{3},\frac{7-\sqrt{13}}{6},\frac{2}{3}$$

$$\therefore$$ $${\int }_0^{\frac{2}{3}}\left[-3x^2+7x-1\right]dx$$

$$={\int }_0^{\frac{7-\sqrt{37}}{6}}(-1)dx+{\int }_{\frac{7-\sqrt{37}}{6}}^{\frac{1}{3}}0dx+{\int }_{\frac{1}{3}}^{\frac{7-\sqrt{13}}{6}}(1)dx+{\int }_{\frac{7-\sqrt{13}}{6}}^{\frac{2}{3}}2dx$$

$$=\frac{-7+\sqrt{37}}{6}+0+\frac{7-\sqrt{13}}{6}-\frac{1}{3}+2\left[\frac{2}{3}-\frac{7-\sqrt{13}}{6}\right]$$

$$=\frac{-7+\sqrt{37}+7-\sqrt{13}-2+8-14+2\sqrt{13}}{6}$$

$$=\frac{\sqrt{37}+\sqrt{13}-8}{6}$$

Now, $${\int }_{\frac{2}{3}}^1\left[3x^2-3x+3\right]dx$$

Let $$f(x)=3x^2-3x+3$$

$$f^{\prime }(x)=6x-3=3(2x-1)$$

$$f^{\prime }(x)=0\Rightarrow 2x-1=0\Rightarrow x=\frac{1}{2}$$

$$f\left(\frac{2}{3}\right)=3\left(\frac{4}{9}-\frac{2}{3}+1\right)$$

$$=\frac{7}{3}=2.3$$

$$f(1)=3(1-1+1)$$

$$=3$$

No integer values present in between $$\frac{7}{3}$$ and 3.

$$\therefore$$ Possible value of $$x=\frac{2}{3},1$$

So, integration don't break anywhere.

$$\therefore$$ $${\int }_{\frac{2}{3}}^1\left[3x^2-3x+3\right]dx$$

$$={\int }_{\frac{2}{3}}^1\left[2.33\right]dx$$

$$={\int }_{\frac{2}{3}}^{1}2dx=2\left(1-\frac{2}{3}\right)=\frac{2}{3}$$

$$\therefore$$ Value of Integration

$$=\frac{\sqrt{37}+\sqrt{13}-1}{6}+\frac{2}{3}$$

$$=\frac{\sqrt{37}+\sqrt{13}-8+4}{6}$$

$$=\frac{\sqrt{37}+\sqrt{13}-4}{6}$$