JEE Main 2022 — Mathematics Calculus

$${\int }_{0}^{2}(|2{x}^{2}-3x|+[x-\frac{1}{2}])dx$$∫02​(∣2x2−3x∣+[x−21​])dx

$$={\int }_{0}^{2}|2{x}^{2}-3x|dx+{\int }_{0}^{2}[x-\frac{1}{2}]dx$$=∫02​∣2x2−3x∣dx+∫02​[x−21​]dx

Let $\displaystyle {I}_{1}={\int }_{0}^{2}|2{x}^{2}-3x|dx={\int }_{0}^{\frac{3}{2}}(3x-2{x}^{2})dx+{\int }_{\frac{3}{2}}^{2}(2{x}^{2}-3x)dx$I1​=∫02​∣2x2−3x∣dx=∫023​​(3x−2x2)dx+∫23​2​(2x2−3x)dx

$$={(\frac{3{x}^{2}}{2}-\frac{2{x}^{3}}{3})}_{0}^{\frac{3}{2}}+{(\frac{2{x}^{3}}{3}-\frac{3{x}^{2}}{2})}_{\frac{3}{2}}^{2}$$=(23x2​−32x3​)023​​+(32x3​−23x2​)23​2​

$$=(\frac{27}{8}-\frac{9}{4})+(\frac{8}{3}-6-\frac{9}{4}+\frac{27}{8})=\frac{9}{4}-\frac{10}{3}=\frac{19}{12}$$=(827​−49​)+(38​−6−49​+827​)=49​−310​=1219​

Let $\displaystyle {I}_{2}={\int }_{0}^{2}[x-\frac{1}{2}]dx$I2​=∫02​[x−21​]dx

Assume $x-\frac{1}{2}=t$x−21​=t

i.e. $\displaystyle {I}_{2}={\int }_{\frac{-1}{2}}^{\frac{3}{2}}[t]dt$I2​=∫2−1​23​​[t]dt

$$={\int }_{-\frac{1}{2}}^{0}(-1)dt+{\int }_{0}^{1}0\cdot dt+{\int }_{1}^{\frac{3}{2}}1\cdot dt$$=∫−21​0​(−1)dt+∫01​0⋅dt+∫123​​1⋅dt

$$={(-t)}_{-\frac{1}{2}}^{0}+{(t)}_{1}^{\frac{3}{2}}=-\frac{1}{2}+(\frac{3}{2}-1)=0$$=(−t)−21​0​+(t)123​​=−21​+(23​−1)=0

Hence, $\displaystyle {\int }_{0}^{2}(|2{x}^{2}-3x|+[x-\frac{1}{2}])dx=\frac{19}{12}$∫02​(∣2x2−3x∣+[x−21​])dx=1219​