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Definite Integration question

2023 · 24 Jan · Shift 1 · Q42
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  5. /2023 · 24 Jan · Shift 1 · Q42

Definite Integration question

2023 · 24 Jan · Shift 1 · Q42

JEE MainMathematicsDefinite IntegrationNumerical+4 / −1
The value of 12∫03∣x2−3x+2∣dx12\int\limits_0^3 {\left| {{x^2} - 3x + 2} \right|dx}120∫3​​x2−3x+2​dx is ‾\underline{\hspace{2cm}}​
Numerical answer
View written solutionFree

Correct answer: 22

  1. We need to evaluate 12∫03∣x2−3x+2∣ dx.12\int_0^3 |x^2-3x+2|\,dx.12∫03​∣x2−3x+2∣dx.

  2. First factor the quadratic: x2−3x+2=(x−1)(x−2).x^2-3x+2=(x-1)(x-2).x2−3x+2=(x−1)(x−2).

So the sign changes at x=1x=1x=1 and x=2x=2x=2.

  1. Determine where the expression is positive/negative:
  • For 0≤x<10\le x<10≤x<1, (x−1)(x−2)>0(x-1)(x-2)>0(x−1)(x−2)>0
  • For 1<x<21<x<21<x<2, (x−1)(x−2)<0(x-1)(x-2)<0(x−1)(x−2)<0
  • For 2<x≤32<x\le 32<x≤3, (x−1)(x−2)>0(x-1)(x-2)>0(x−1)(x−2)>0

Hence,

\begin{cases} x^2-3x+2, & 0\le x\le 1,\\ -(x^2-3x+2), & 1\le x\le 2,\\ x^2-3x+2, & 2\le x\le 3. \end{cases}$$ 4. Therefore, $$\int_0^3 |x^2-3x+2|dx =\int_0^1 (x^2-3x+2)dx-\int_1^2 (x^2-3x+2)dx+\int_2^3 (x^2-3x+2)dx.$$ 5. Antiderivative of $x^2-3x+2$ is $$\int (x^2-3x+2)dx=\frac{x^3}{3}-\frac{3x^2}{2}+2x.$$ Let $$F(x)=\frac{x^3}{3}-\frac{3x^2}{2}+2x.$$ 6. Compute each part: For $[0,1]$: $$\int_0^1 (x^2-3x+2)dx=F(1)-F(0).$$ Now, $$F(1)=\frac13-\frac32+2=\frac{2-9+12}{6}=\frac56,$$ $$F(0)=0.$$ So, $$\int_0^1 (x^2-3x+2)dx=\frac56.$$ For $[1,2]$: $$\int_1^2 (x^2-3x+2)dx=F(2)-F(1).$$ Now, $$F(2)=\frac{8}{3}-6+4=\frac{8}{3}-2=\frac23.$$ Thus, $$\int_1^2 (x^2-3x+2)dx=\frac23-\frac56=\frac{4-5}{6}=-\frac16.$$ Hence, $$\int_1^2 |x^2-3x+2|dx=\frac16.$$ For $[2,3]$: $$\int_2^3 (x^2-3x+2)dx=F(3)-F(2).$$ Now, $$F(3)=9-\frac{27}{2}+6=15-\frac{27}{2}=\frac32.$$ So, $$\int_2^3 (x^2-3x+2)dx=\frac32-\frac23=\frac{9-4}{6}=\frac56.$$ 7. Add them: $$\int_0^3 |x^2-3x+2|dx=\frac56+\frac16+\frac56=\frac{11}{6}.$$ 8. Multiply by $12$: $$12\int_0^3 |x^2-3x+2|dx=12\cdot \frac{11}{6}=22.$$ Therefore, the required integer is $$\boxed{22}.$$
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