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Integral Calculus: JEE Main Mathematics Question with Solution

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

Because when and when . The roots of are and , which are where the expression changes sign inside .

The value of

12 \int_0^3 |x^2 - 3x + 2|dx

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Correct answer

The value of $12 \int_0^3 |x^2 - 3x + 2|dx$ is $22$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

The integral expression is $I = 12 \int_0^3 |x^2 - 3x + 2| \, dx$ .

02goal

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

03approach

Factor the quadratic $x^2 - 3x + 2 = (x - 1)(x - 2)$ . 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: $\int (x^2 - 3x + 2) \, dx = \frac{x^3}{3} - \frac{3x^2}{2} + 2x$ . On $[0, 1]$ : $\left[\frac{1}{3} - \frac{3}{2} + 2\right] - 0 = \frac{5}{6}$ . On $[1, 2]$ : the quadratic is negative, so $-\int_1^2 (x^2 - 3x + 2) \, dx = -\left[\left(\frac{8}{3} - 6 + 4\right) - \left(\frac{1}{3} - \frac{3}{2} + 2\right)\right] = -\left[\frac{2}{3} - \frac{5}{6}\right] = \frac{1}{6}$ . On $[2, 3]$ : $\left[\left(9 - \frac{27}{2} + 6\right) - \left(\frac{8}{3} - 6 + 4\right)\right] = \frac{3}{2} - \frac{2}{3} = \frac{5}{6}$ . 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 $y = x^2 - 3x + 2$ 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 $x^2 - 3x + 2$ 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 $\int_0^3 |x^2 - 3x + 2|\,dx$ split across the intervals of constant sign?
$\int_0^1 (x^2 - 3x + 2)\,dx - \int_1^2 (x^2 - 3x + 2)\,dx + \int_2^3 (x^2 - 3x + 2)\,dx$
What is the antiderivative $F(x) = \int (x^2 - 3x + 2)\,dx$ with $F(0) = 0$ ?
$F(x) = \frac{x^3}{3} - \frac{3x^2}{2} + 2x$
Evaluating $F(1) = \frac{5}{6}$ , $F(2) = \frac{2}{3}$ , and $F(3) = \frac{3}{2}$ , what is the final value of $12 \int_0^3 |x^2 - 3x + 2|\,dx$ ?
$22$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $f(x) \ge 0$ 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 $[0, 3]$ .

Answer

The value of $12 \int_0^3 |x^2 - 3x + 2|dx$ is $22$ .

Why each option works or fails

Step-by-step solution

given: The integral expression is $I = 12 \int_0^3 |x^2 - 3x + 2| \, dx$ .
goal: Evaluate the definite integral and determine the integer value of $I$ .
approach: Factor the quadratic $x^2 - 3x + 2 = (x - 1)(x - 2)$ . 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: $\int (x^2 - 3x + 2) \, dx = \frac{x^3}{3} - \frac{3x^2}{2} + 2x$ . On $[0, 1]$ : $\left[\frac{1}{3} - \frac{3}{2} + 2\right] - 0 = \frac{5}{6}$ . On $[1, 2]$ : the quadratic is negative, so $-\int_1^2 (x^2 - 3x + 2) \, dx = -\left[\left(\frac{8}{3} - 6 + 4\right) - \left(\frac{1}{3} - \frac{3}{2} + 2\right)\right] = -\left[\frac{2}{3} - \frac{5}{6}\right] = \frac{1}{6}$ . On $[2, 3]$ : $\left[\left(9 - \frac{27}{2} + 6\right) - \left(\frac{8}{3} - 6 + 4\right)\right] = \frac{3}{2} - \frac{2}{3} = \frac{5}{6}$ . 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 $y = x^2 - 3x + 2$ 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.

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

The value of $12 \int_0^3 |x^2 - 3x + 2|dx$ is $22$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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Integral Calculus: JEE Main Mathematics Question with Solution

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