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Definite Integration question

2022 · 27 Jul · Shift 2 · Q29

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Mathematics
Definite Integration
2022 · 27 Jul · Shift 2 · Q29

Definite Integration question

2022 · 27 Jul · Shift 2 · Q29

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1

\int\limits_{0}^{2}\left(\left|2 x^{2}-3 x\right|+\left[x-\frac{1}{2}\right]\right) \mathrm{d} x

, where [t] is the greatest integer function, is equal to :

A
$\frac{7}{6}$
B
$\frac{19}{12}$
C
$\frac{31}{12}$
D
$\frac{3}{2}$

View written solutionFree

Correct answer: B

We need to evaluate $I=\int_0^2\left(|2x^2-3x|+\left[x-\frac12\right]\right)dx.$ So split it as $I=\int_0^2 |2x^2-3x|\,dx+\int_0^2 \left[x-\frac12\right]dx.$
First part: evaluate $I_1=\int_0^2 |2x^2-3x|\,dx.$ Factor the expression inside modulus: $2x^2-3x=x(2x-3).$ Its zeros are at $x=0,\quad x=\frac32.$

Now check sign on $[0,2]$ :

For $0<x<\frac32$ , we have $x>0$ and $2x-3<0$ , so $2x^2-3x<0$ .
For $\frac32<x\le 2$ , we have $2x^2-3x>0$ .

Hence

\begin{cases} -(2x^2-3x)=3x-2x^2, & 0\le x\le \frac32,\\[4pt] 2x^2-3x, & \frac32\le x\le 2. \end{cases}$$ Therefore, $$I_1=\int_0^{3/2}(3x-2x^2)dx+\int_{3/2}^2(2x^2-3x)dx.$$ Compute the first integral: $$\int (3x-2x^2)dx=\frac{3x^2}{2}-\frac{2x^3}{3}.$$ So $$\int_0^{3/2}(3x-2x^2)dx =\left[\frac{3x^2}{2}-\frac{2x^3}{3}\right]_0^{3/2} =\frac{3}{2}\cdot\frac{9}{4}-\frac{2}{3}\cdot\frac{27}{8} =\frac{27}{8}-\frac{9}{4} =\frac{9}{8}.$$ Compute the second integral: $$\int (2x^2-3x)dx=\frac{2x^3}{3}-\frac{3x^2}{2}.$$ So $$\int_{3/2}^2(2x^2-3x)dx =\left[\frac{2x^3}{3}-\frac{3x^2}{2}\right]_{3/2}^2.$$ At $x=2$: $$\frac{2(8)}{3}-\frac{3(4)}{2}=\frac{16}{3}-6=-\frac{2}{3}.$$ At $x=\frac32$: $$\frac{2}{3}\cdot\frac{27}{8}-\frac{3}{2}\cdot\frac{9}{4}=\frac{9}{4}-\frac{27}{8}=-\frac{9}{8}.$$ Thus, $$\int_{3/2}^2(2x^2-3x)dx=-\frac{2}{3}-\left(-\frac{9}{8}\right)=\frac{11}{24}.$$ So, $$I_1=\frac98+\frac{11}{24}=\frac{27}{24}+\frac{11}{24}=\frac{38}{24}=\frac{19}{12}.$$ 3. Second part: evaluate $$I_2=\int_0^2 \left[x-\frac12\right]dx.$$ We analyze the greatest integer function. For $0\le x<\frac12$: $$0\le x-\frac12<0 \implies \left[x-\frac12\right]=-1.$$ For $\frac12\le x<\frac32$: $$0\le x-\frac12<1 \implies \left[x-\frac12\right]=0.$$ For $\frac32\le x<2$: $$1\le x-\frac12<\frac32 \implies \left[x-\frac12\right]=1.$$ (The value at isolated points does not affect the integral.) Thus, $$I_2=\int_0^{1/2}(-1)dx+\int_{1/2}^{3/2}0\,dx+\int_{3/2}^2 1\,dx =-\frac12+0+\frac12=0.$$ 4. Therefore, $$I=I_1+I_2=\frac{19}{12}+0=\frac{19}{12}.$$ 5. Comparing with the options: - A: $\frac76$ - B: $\frac{19}{12}$ - C: $\frac{31}{12}$ - D: $\frac32$ So the correct option is **B**.

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