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Integral Calculus: JEE Main Mathematics Question with Solution

Q: Why does substituting not change the value of the area?

A substitution is a rigid horizontal translation of the entire region, and rigid translations preserve area ( ).

The area of the region

\{(\mathrm{x},\mathrm{y}):\mathrm{x}^{2}+4\mathrm{x}+2\le \mathrm{y}\le |\mathrm{x}+2|\}

is equal to

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7

24/5

20/3

5

Step-by-step solutionView

Correct answer

By shifting coordinates with $t = x+2$ , the region becomes symmetric with bounding curves $y = t^2 - 2$ and $y = |t|$ , yielding an area of $20/3$ .

Option analysis

Why each option works or fails

A ·

7

Integrating only one half of the symmetric region and incorrectly doubling or miscalculating $\int_0^2 (|t| - (t^2 - 2))\,dt$ as $7/2$ . Evaluate $\int_0^2 (t - t^2 + 2)\,dt = [\frac{t^2}{2} - \frac{t^3}{3} + 2t]_0^2 = 2 - \frac{8}{3} + 4 = \frac{10}{3}$ , then double it to get $20/3$ .

B ·

24/5

Using an incorrect exponent or integration rule for the polynomial terms, leading to a denominator of 5. Integrate powers using $\int t^n\,dt = \frac{t^{n+1}}{n+1}$ , which yields denominators of 2 and 3, not 5.

C ·

20/3

This is the correct option. The curves intersect at $x+2 = \pm 2$ , and integrating the upper curve $|x+2|$ minus the lower parabola $(x+2)^2 - 2$ across $[-4, 0]$ gives $\frac{20}{3}$ .

D ·

5

Calculating the area without doubling the symmetric half, or dropping the linear term during integration. Ensure both symmetric sides on $t \in [-2, 2]$ are included, giving $2 \times \frac{10}{3} = \frac{20}{3}$ .

Reviewed route

Solution

StepWorking

01given

Region bounded by $y \ge x^2 + 4x + 2$ and $y \le |x + 2|$ .

02approach

Rewrite the parabola as $y = (x+2)^2 - 2$ . By shifting the origin horizontally using $X = x + 2$ , the region is bounded by $Y = X^2 - 2$ and $Y = |X|$ , which preserves the area and provides symmetry about the Y-axis.

03execute

Find the points of intersection: $|X|^2 - |X| - 2 = 0 \implies (|X|-2)(|X|+1) = 0 \implies |X| = 2 \implies X = \pm 2$ .

04execute

Set up the area integral using even symmetry: $\text{Area} = 2 \int_{0}^{2} \left(X - (X^2 - 2)\right) dX = 2 \left[ \frac{X^2}{2} - \frac{X^3}{3} + 2X \right]_0^2 = 2 \left( 2 - \frac{8}{3} + 4 \right) = 2 \left( \frac{10}{3} \right) = \frac{20}{3}$ .

✓verify

The enclosing triangle with base from $X=-2$ to $X=2$ at $Y=2$ has area $\frac{1}{2} \times 4 \times 2 = 4$ , and the region below $Y=0$ between the parabola and Y=0 adds $\approx 2 \times \int_0^{\sqrt{2}} (2 - X^2) dX = \frac{8\sqrt{2}}{3} \approx 3.77$ . Total area $\approx 4 + 3.77 - \text{overlap} = 6.67 = 20/3$ , which is consistent.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why does substituting

X = x+2

not change the value of the area?

A substitution $X = x + c$ is a rigid horizontal translation of the entire region, and rigid translations preserve area ( $dX = dx$ ).

Answer

By shifting coordinates with $t = x+2$ , the region becomes symmetric with bounding curves $y = t^2 - 2$ and $y = |t|$ , yielding an area of $20/3$ .

Why each option works or fails

A: $7$ - Integrating only one half of the symmetric region and incorrectly doubling or miscalculating $\int_0^2 (|t| - (t^2 - 2))\,dt$ as $7/2$ . Evaluate $\int_0^2 (t - t^2 + 2)\,dt = [\frac{t^2}{2} - \frac{t^3}{3} + 2t]_0^2 = 2 - \frac{8}{3} + 4 = \frac{10}{3}$ , then double it to get $20/3$ .
B: $24/5$ - Using an incorrect exponent or integration rule for the polynomial terms, leading to a denominator of 5. Integrate powers using $\int t^n\,dt = \frac{t^{n+1}}{n+1}$ , which yields denominators of 2 and 3, not 5.
C · correct: $20/3$ - This is the correct option. The curves intersect at $x+2 = \pm 2$ , and integrating the upper curve $|x+2|$ minus the lower parabola $(x+2)^2 - 2$ across $[-4, 0]$ gives $\frac{20}{3}$ .
D: $5$ - Calculating the area without doubling the symmetric half, or dropping the linear term during integration. Ensure both symmetric sides on $t \in [-2, 2]$ are included, giving $2 \times \frac{10}{3} = \frac{20}{3}$ .

Step-by-step solution

given: Region bounded by $y \ge x^2 + 4x + 2$ and $y \le |x + 2|$ .
approach: Rewrite the parabola as $y = (x+2)^2 - 2$ . By shifting the origin horizontally using $X = x + 2$ , the region is bounded by $Y = X^2 - 2$ and $Y = |X|$ , which preserves the area and provides symmetry about the Y-axis.
execute: Find the points of intersection: $|X|^2 - |X| - 2 = 0 \implies (|X|-2)(|X|+1) = 0 \implies |X| = 2 \implies X = \pm 2$ .
execute: Set up the area integral using even symmetry: $\text{Area} = 2 \int_{0}^{2} \left(X - (X^2 - 2)\right) dX = 2 \left[ \frac{X^2}{2} - \frac{X^3}{3} + 2X \right]_0^2 = 2 \left( 2 - \frac{8}{3} + 4 \right) = 2 \left( \frac{10}{3} \right) = \frac{20}{3}$ .
verify: The enclosing triangle with base from $X=-2$ to $X=2$ at $Y=2$ has area $\frac{1}{2} \times 4 \times 2 = 4$ , and the region below $Y=0$ between the parabola and Y=0 adds $\approx 2 \times \int_0^{\sqrt{2}} (2 - X^2) dX = \frac{8\sqrt{2}}{3} \approx 3.77$ . Total area $\approx 4 + 3.77 - \text{overlap} = 6.67 = 20/3$ , which is consistent.

Integral Calculus: JEE Main Mathematics Question with Solution

By shifting coordinates with t=x+2t = x+2t=x+2, the region becomes symmetric with bounding curves y=t2−2y = t^2 - 2y=t2−2 and y=∣t∣y = |t|y=∣t∣, yielding an area of 20/320/320/3.

01given

02approach

03execute

04execute

✓verify

Students also ask

By shifting coordinates with $t = x+2$ , the region becomes symmetric with bounding curves $y = t^2 - 2$ and $y = |t|$ , yielding an area of $20/3$ .