JEE MainMathematics

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Definite Integral of an Absolute Value Quadratic Function

Q: Why does the integral split at x = 1 and x = 2?

Because |f(x)| = f(x) when f(x) ≥ 0 and |f(x)| = -f(x) when f(x) < 0. The roots of x^2 - 3x + 2 = 0 are x = 1 and x = 2, which are where the expression changes sign inside [0, 3].

The value of 12 integral from 0 to 3 of |x^2 - 3x + 2|dx is

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Correct answer

The value of $12 \int_{0}^{3} ∣ x^{2} - 3 x + 2 ∣ d x$ is $22$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

The integral expression is $I = 12 \int_{0}^{3} ∣ x^{2} - 3 x + 2 ∣ d x$ .

02Goal

Evaluate the definite integral and determine the integer value of $I$ .

03Approach

Factor the quadratic $x^{2} - 3 x + 2 = (x - 1) (x - 2)$ . Determine its sign over $[0, 3]$ : positive on $[0, 1)$ and $(2, 3]$ , negative on $(1, 2)$ . Split the integral into three sub-intervals: $[0, 1]$ , $[1, 2]$ , and $[2, 3]$ .

04Execute

Evaluate each piece: $\int (x^{2} - 3 x + 2) d x = \frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x$ . On $[0, 1]$ : $[\frac{1}{3} - \frac{3}{2} + 2] - 0 = \frac{5}{6}$ . On $[1, 2]$ : the quadratic is negative, so $- \int_{1}^{2} (x^{2} - 3 x + 2) d x = - [(\frac{8}{3} - 6 + 4) - (\frac{1}{3} - \frac{3}{2} + 2)] = - [\frac{2}{3} - \frac{5}{6}] = \frac{1}{6}$ . On $[2, 3]$ : $[(9 - \frac{27}{2} + 6) - (\frac{8}{3} - 6 + 4)] = \frac{3}{2} - \frac{2}{3} = \frac{5}{6}$ . Sum of integrals: $\frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{11}{6}$ . Multiply by $12$ : $12 \times \frac{11}{6} = 22$ .

✓Verify

By symmetry of the parabola $y = x^{2} - 3 x + 2$ about its vertex at $x = 1.5$ , the area from $0$ to $1$ must equal the area from $2$ to $3$ (both equal $5 / 6$ ). The area between the roots $[1, 2]$ is given by Archimedes' formula $\frac{∣ a ∣}{6} (x_{2} - x_{1})^{3} = \frac{1}{6} (1)^{3} = \frac{1}{6}$ . Total integral $= 5 / 6 + 1 / 6 + 5 / 6 = 11 / 6$ . Multiplying by 12 gives $22$ . The result is completely verified.

Hints that build this answer step by step

Where does the integrand $x^{2} - 3 x + 2$ change sign on the interval $[0, 3]$ ?
At $x = 1$ and $x = 2$ , where it is negative on $(1, 2)$ and positive on $[0, 1)$ and $(2, 3]$ .
How is the integral $\int_{0}^{3} ∣ x^{2} - 3 x + 2 ∣ d x$ split across the intervals of constant sign?
$\int_{0}^{1} (x^{2} - 3 x + 2) d x - \int_{1}^{2} (x^{2} - 3 x + 2) d x + \int_{2}^{3} (x^{2} - 3 x + 2) d x$
What is the antiderivative $F (x) = \int (x^{2} - 3 x + 2) d x$ with $F (0) = 0$ ?
$F (x) = \frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x$
Evaluating $F (1) = \frac{5}{6}$ , $F (2) = \frac{2}{3}$ , and $F (3) = \frac{3}{2}$ , what is the final value of $12 \int_{0}^{3} ∣ x^{2} - 3 x + 2 ∣ d x$ ?
$22$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why does the integral split at x = 1 and x = 2?

Because $∣ f (x) ∣ = f (x)$ when $f (x) \geq 0$ and $∣ f (x) ∣ = - f (x)$ when $f (x) < 0$ . The roots of $x^{2} - 3 x + 2 = 0$ are $x = 1$ and $x = 2$ , which are where the expression changes sign inside $[0, 3]$ .

Answer

The value of $12 \int_{0}^{3} ∣ x^{2} - 3 x + 2 ∣ d x$ is $22$ .

Why each option works or fails

Step-by-step solution

Given: The integral expression is $I = 12 \int_{0}^{3} ∣ x^{2} - 3 x + 2 ∣ d x$ .
Goal: Evaluate the definite integral and determine the integer value of $I$ .
Approach: Factor the quadratic $x^{2} - 3 x + 2 = (x - 1) (x - 2)$ . Determine its sign over $[0, 3]$ : positive on $[0, 1)$ and $(2, 3]$ , negative on $(1, 2)$ . Split the integral into three sub-intervals: $[0, 1]$ , $[1, 2]$ , and $[2, 3]$ .
Execute: Evaluate each piece: $\int (x^{2} - 3 x + 2) d x = \frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x$ . On $[0, 1]$ : $[\frac{1}{3} - \frac{3}{2} + 2] - 0 = \frac{5}{6}$ . On $[1, 2]$ : the quadratic is negative, so $- \int_{1}^{2} (x^{2} - 3 x + 2) d x = - [(\frac{8}{3} - 6 + 4) - (\frac{1}{3} - \frac{3}{2} + 2)] = - [\frac{2}{3} - \frac{5}{6}] = \frac{1}{6}$ . On $[2, 3]$ : $[(9 - \frac{27}{2} + 6) - (\frac{8}{3} - 6 + 4)] = \frac{3}{2} - \frac{2}{3} = \frac{5}{6}$ . Sum of integrals: $\frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{11}{6}$ . Multiply by $12$ : $12 \times \frac{11}{6} = 22$ .
Verify: By symmetry of the parabola $y = x^{2} - 3 x + 2$ about its vertex at $x = 1.5$ , the area from $0$ to $1$ must equal the area from $2$ to $3$ (both equal $5 / 6$ ). The area between the roots $[1, 2]$ is given by Archimedes' formula $\frac{∣ a ∣}{6} (x_{2} - x_{1})^{3} = \frac{1}{6} (1)^{3} = \frac{1}{6}$ . Total integral $= 5 / 6 + 1 / 6 + 5 / 6 = 11 / 6$ . Multiplying by 12 gives $22$ . The result is completely verified.

Definite Integral of an Absolute Value Quadratic Function

What feels right?

The value of $12 \int_{0}^{3} ∣ x^{2} - 3 x + 2 ∣ d x$ is $22$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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