JEE Main 2022 — Mathematics Calculus

$$I={\int }_{0}^{1}[2x-|3{x}^{2}-3x-2x+2|+1]dx$$I=∫01​[2x−∣3x2−3x−2x+2∣+1]dx

$$I={\int }_{0}^{1}[2x-|(3x-2)(x-1)|]dx+{\int }_{0}^{1}1dx$$I=∫01​[2x−∣(3x−2)(x−1)∣]dx+∫01​1dx

$$I={\int }_{0}^{\frac{2}{3}}[(2x-(3{x}^{2}-5x+2))]dx+{\int }_{\frac{2}{3}}^{1}(2x+(3{x}^{2}-5x+2))dx+1$$I=∫032​​[(2x−(3x2−5x+2))]dx+∫32​1​(2x+(3x2−5x+2))dx+1

$$I={\int }_{0}^{\frac{2}{3}}[-3{x}^{2}+7x-2]dx+{\int }_{\frac{2}{3}}^{1}(3{x}^{2}-3x+2)dx+1$$I=∫032​​[−3x2+7x−2]dx+∫32​1​(3x2−3x+2)dx+1

Let $\displaystyle {I}_{1}={\int }_{0}^{\frac{2}{3}}[-3{x}^{2}+7x-2]dx$I1​=∫032​​[−3x2+7x−2]dx

The graph of $y=-3{x}^{2}+7x-2$y=−3x2+7x−2 and

$${I}_{2}={\int }_{\frac{2}{3}}^{1}(3{x}^{2}-3x+2)dx$$I2​=∫32​1​(3x2−3x+2)dx

So $\displaystyle {I}_{1}={\int }_{0}^{\alpha }(-2)dx+{\int }_{\alpha }^{\frac{1}{3}}(-1)dx+{\int }_{1/3}^{\beta }0dx+{\int }_{\beta }^{\frac{2}{3}}1\cdot dx$I1​=∫0α​(−2)dx+∫α31​​(−1)dx+∫1/3β​0dx+∫β32​​1⋅dx

$$=-2\alpha -(\frac{1}{3}-\alpha )+\frac{2}{3}-\beta =-\alpha -\beta +\frac{1}{3}$$=−2α−(31​−α)+32​−β=−α−β+31​

$$y=3{x}^{2}-3x+2$$y=3x2−3x+2

When $x\in (\frac{2}{3},1)$x∈(32​,1)

$$3{x}^{2}-3x+2\in (\frac{4}{3},2)$$3x2−3x+2∈(34​,2)

$$[3{x}^{2}-3x+2]=1$$[3x2−3x+2]=1

$$\therefore {I}_{2}={\int }_{\frac{2}{3}}^{1}[3{x}^{2}-3x+2]dx=1(1-\frac{2}{3})=\frac{1}{3}$$∴I2​=∫32​1​[3x2−3x+2]dx=1(1−32​)=31​

Hence $I=(\frac{1}{3}-(\alpha +\beta ))+(\frac{1}{3})+1$I=(31​−(α+β))+(31​)+1

$$=\frac{5}{3}-(\frac{7-\sqrt{37}}{6}+\frac{7-\sqrt{13}}{6})$$=35​−(67−37​​+67−13​​)

$=\frac{-2}{3}+\frac{\sqrt{37}+\sqrt{13}}{6}$=3−2​+637​+13​​$=\frac{-2}{3}+\frac{\sqrt{37}+\sqrt{13}}{6}$=3−2​+637​+13​​

$$=\frac{\sqrt{37}+\sqrt{13}-4}{6}$$=637​+13​−4​