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Integral Calculus: JEE Main Mathematics Question with Solution

Q: Does this formula work for any line and parabola?

Yes, for any parabola and any line intersecting it at two points with y-coordinates and , the enclosed area is always .

The area enclosed by the curves

y^2 + 4x = 4

and

y - 2x = 2

is :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

9

\frac{22}{3}

\frac{23}{3}

\frac{25}{3}

Step-by-step solutionView

Correct answer

The area enclosed between the parabola $x = 1 - \frac{y^2}{4}$ and the line $x = \frac{y - 2}{2}$ is $9$ .

Option analysis

Why each option works or fails

A ·

9

None. This option correctly evaluates the enclosed area. Integrating with respect to $y$ from $y = -4$ to $y = 2$ gives $\int_{-4}^2 \left(2 - \frac{y}{2} - \frac{y^2}{4}\right) dy = 9$ .

B ·

\frac{22}{3}

Evaluating the integral with an arithmetic sign error in the constant term or using asymmetric integration limits leads to $\frac{22}{3}$ . Ensure the bounds are found by equating $4 - y^2 = 2y - 4$ , giving $y^2 + 2y - 8 = 0$ , so $y = -4$ and $y = 2$ .

C ·

\frac{23}{3}

Making a slip when evaluating the cubic term $\left[-\frac{y^3}{12}\right]_{-4}^2$ gives a net total of $\frac{23}{3}$ . Check that $\left[-\frac{y^3}{12}\right]_{-4}^2 = -\frac{8}{12} - \left(-\frac{-64}{12}\right) = -\frac{8}{12} - \frac{64}{12} = -6$ .

D ·

\frac{25}{3}

Adding an extra $\frac{2}{3}$ or making a sign slip when substituting the lower limit $y = -4$ yields $\frac{25}{3}$ . Carefully compute each term: $[2y]_{-4}^2 = 12$ , $\left[-\frac{y^2}{4}\right]_{-4}^2 = -3$ , and $\left[-\frac{y^3}{12}\right]_{-4}^2 = -6$ , summing to $12 - 3 - 6 = 9$ .

Reviewed route

Solution

StepWorking

01given

The boundary curves are the parabola $y^2 + 4x = 4$ and the straight line $y - 2x = 2$ .

02goal

Find the area of the region enclosed between the parabola and the line.

03approach

Express $x$ as a function of $y$ for both curves, find their intersection points in terms of $y$ , and integrate $(x_{\text{right}} - x_{\text{left}})\,dy$ from the lower to the upper $y$ -limit.

04execute

From the line, $2x = y - 2 \implies x = \frac{y-2}{2}$ . Substitute into the parabola: $y^2 + 4\left(\frac{y-2}{2}\right) = 4 \implies y^2 + 2y - 4 = 4 \implies y^2 + 2y - 8 = 0$ . Factoring gives $(y+4)(y-2) = 0$ , so $y = -4$ and $y = 2$ .

05execute

Set up the area integral with $x_{\text{right}} = \frac{4-y^2}{4}$ and $x_{\text{left}} = \frac{y-2}{2}$ : $A = \int_{-4}^{2} \left(\frac{4-y^2}{4} - \frac{y-2}{2}\right) dy = \int_{-4}^{2} \left(2 - \frac{y}{2} - \frac{y^2}{4}\right) dy$ . Evaluating the antiderivative: $\left[ 2y - \frac{y^2}{4} - \frac{y^3}{12} \right]_{-4}^{2} = \left(4 - 1 - \frac{8}{12}\right) - \left(-8 - 4 - \frac{-64}{12}\right) = \frac{7}{3} - \left(-12 + \frac{16}{3}\right) = \frac{7}{3} - \left(-\frac{20}{3}\right) = 9$ .

✓verify

Using Archimedes' formula for a parabolic segment cut by a chord: $\text{Area} = \frac{|a|}{6}(y_2 - y_1)^3$ for a parabola $x = -\frac{1}{4}y^2 + 1$ . Here $|a| = \frac{1}{4}$ , $y_2 - y_1 = 2 - (-4) = 6$ . So $\text{Area} = \frac{1/4}{6} \times 6^3 = \frac{1}{24} \times 216 = 9$ . The result is verified.

✓ 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: 9 September 2026

Quick checks

Students also ask

Why integrate with respect to y instead of x?

Integrating with respect to x requires splitting the region into two parts because the upper and lower boundary curves change at x = 0. Integrating with respect to y requires just a single integral.

Does this formula work for any line and parabola?

Yes, for any parabola $x = Ay^2 + By + C$ and any line intersecting it at two points with y-coordinates $y_1$ and $y_2$ , the enclosed area is always $\frac{|A|}{6}|y_2 - y_1|^3$ .

Answer

The area enclosed between the parabola $x = 1 - \frac{y^2}{4}$ and the line $x = \frac{y - 2}{2}$ is $9$ .

Why each option works or fails

A · correct: $9$ - None. This option correctly evaluates the enclosed area. Integrating with respect to $y$ from $y = -4$ to $y = 2$ gives $\int_{-4}^2 \left(2 - \frac{y}{2} - \frac{y^2}{4}\right) dy = 9$ .
B: $\frac{22}{3}$ - Evaluating the integral with an arithmetic sign error in the constant term or using asymmetric integration limits leads to $\frac{22}{3}$ . Ensure the bounds are found by equating $4 - y^2 = 2y - 4$ , giving $y^2 + 2y - 8 = 0$ , so $y = -4$ and $y = 2$ .
C: $\frac{23}{3}$ - Making a slip when evaluating the cubic term $\left[-\frac{y^3}{12}\right]_{-4}^2$ gives a net total of $\frac{23}{3}$ . Check that $\left[-\frac{y^3}{12}\right]_{-4}^2 = -\frac{8}{12} - \left(-\frac{-64}{12}\right) = -\frac{8}{12} - \frac{64}{12} = -6$ .
D: $\frac{25}{3}$ - Adding an extra $\frac{2}{3}$ or making a sign slip when substituting the lower limit $y = -4$ yields $\frac{25}{3}$ . Carefully compute each term: $[2y]_{-4}^2 = 12$ , $\left[-\frac{y^2}{4}\right]_{-4}^2 = -3$ , and $\left[-\frac{y^3}{12}\right]_{-4}^2 = -6$ , summing to $12 - 3 - 6 = 9$ .

Step-by-step solution

given: The boundary curves are the parabola $y^2 + 4x = 4$ and the straight line $y - 2x = 2$ .
goal: Find the area of the region enclosed between the parabola and the line.
approach: Express $x$ as a function of $y$ for both curves, find their intersection points in terms of $y$ , and integrate $(x_{\text{right}} - x_{\text{left}})\,dy$ from the lower to the upper $y$ -limit.
execute: From the line, $2x = y - 2 \implies x = \frac{y-2}{2}$ . Substitute into the parabola: $y^2 + 4\left(\frac{y-2}{2}\right) = 4 \implies y^2 + 2y - 4 = 4 \implies y^2 + 2y - 8 = 0$ . Factoring gives $(y+4)(y-2) = 0$ , so $y = -4$ and $y = 2$ .
execute: Set up the area integral with $x_{\text{right}} = \frac{4-y^2}{4}$ and $x_{\text{left}} = \frac{y-2}{2}$ : $A = \int_{-4}^{2} \left(\frac{4-y^2}{4} - \frac{y-2}{2}\right) dy = \int_{-4}^{2} \left(2 - \frac{y}{2} - \frac{y^2}{4}\right) dy$ . Evaluating the antiderivative: $\left[ 2y - \frac{y^2}{4} - \frac{y^3}{12} \right]_{-4}^{2} = \left(4 - 1 - \frac{8}{12}\right) - \left(-8 - 4 - \frac{-64}{12}\right) = \frac{7}{3} - \left(-12 + \frac{16}{3}\right) = \frac{7}{3} - \left(-\frac{20}{3}\right) = 9$ .
verify: Using Archimedes' formula for a parabolic segment cut by a chord: $\text{Area} = \frac{|a|}{6}(y_2 - y_1)^3$ for a parabola $x = -\frac{1}{4}y^2 + 1$ . Here $|a| = \frac{1}{4}$ , $y_2 - y_1 = 2 - (-4) = 6$ . So $\text{Area} = \frac{1/4}{6} \times 6^3 = \frac{1}{24} \times 216 = 9$ . The result is verified.

Shortcut:

When to use it: Whenever finding the area enclosed between any parabola and any secant line.

given: Parabola $x = 1 - \frac{1}{4}y^2$ (leading coefficient in $x = -\frac{1}{4}y^2 + 1$ is $A = -1/4$ ) intersected by the chord from $y = -4$ to $y = 2$ .

approach: The area between a parabola $x = Ay^2 + By + C$ and a secant line intersecting it at $y_1, y_2$ is given by $\frac{|A|}{6}(y_2 - y_1)^3$ .

execute: Here $|A| = \frac{1}{4}$ , $y_1 = -4$ , $y_2 = 2$ . The difference is $y_2 - y_1 = 6$ . Therefore, $\text{Area} = \frac{1}{6} \times \frac{1}{4} \times 6^3 = \frac{216}{24} = 9$ .

verify: Since $\frac{216}{24} = 9$ , the computation matches the standard integration method.

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

The area enclosed between the parabola $x = 1 - \frac{y^2}{4}$ and the line $x = \frac{y - 2}{2}$ is $9$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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Integral Calculus: JEE Main Mathematics Question with Solution

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