JEE MainMathematics

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How do you find the area bounded by a parabola and a line?

Q: Is there a direct formula for the area between a parabola and a line?

Yes, for a parabola y = ax^2 + bx + c intersecting a line at x_1 and x_2, the bounded area is simply (|a|)/6(x_2 - x_1)^3.

The area of the region {(x, y): x^2 - 8x ≤ y ≤ -x} is:

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A343/6B637/6C437/6D523/6

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Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

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Is there a direct formula for the area between a parabola and a line?

Yes, for a parabola $y = a x^{2} + b x + c$ intersecting a line at $x_{1}$ and $x_{2}$ , the bounded area is simply $\frac{∣ a ∣}{6} (x_{2} - x_{1})^{3}$ .

Answer

The enclosed region is bounded between $x = 0$ and $x = 7$ , yielding an area of $\frac{343}{6}$ .

Why each option works or fails

A · correct: 343/6 - This is the correct calculation of the bounded area. Integrating $(- x) - (x^{2} - 8 x) = 7 x - x^{2}$ from $x = 0$ to $x = 7$ correctly yields $\frac{7 (49)}{2} - \frac{343}{3} = \frac{343}{6}$ .
B: 637/6 - Adding the terms instead of subtracting when integrating $[\frac{7 x^{2}}{2} - \frac{x^{3}}{3}]$ leads to $\frac{343}{2} + \frac{343}{3} = \frac{1715}{6}$ or similar sign errors combined with numerator arithmetic slips yielding $\frac{637}{6}$ . Ensure you correctly evaluate $\frac{7 (49)}{2} - \frac{343}{3} = 343 (\frac{1}{2} - \frac{1}{3}) = \frac{343}{6}$ .
C: 437/6 - Incorrectly evaluating $7^{3}$ or miscalculating common denominators leads to $\frac{437}{6}$ . Factor out $7^{3} = 343$ before subtracting fractions: $343 \times \frac{1}{6} = \frac{343}{6}$ .
D: 523/6 - Using incorrect intersection bounds or making an arithmetic mistake when evaluating $(7 x - x^{2})$ at the upper limit. Find the roots of $- x = x^{2} - 8 x ⟹ x (x - 7) = 0$ , giving exact limits $0$ and $7$ .

Step-by-step solution

Given: Region is defined by $x^{2} - 8 x \leq y \leq - x$ . The lower boundary curve is $y_{1} = x^{2} - 8 x$ and the upper boundary curve is $y_{2} = - x$ .
Goal: Find the area of the bounded region, $A = \int_{a}^{b} (y_{upper} - y_{lower}) d x$ .
Approach: Find the points of intersection by equating $x^{2} - 8 x = - x$ , determine the integration limits, and compute $\int_{a}^{b} (- x - (x^{2} - 8 x)) d x$ .
Execute: Solve for intersection: $x^{2} - 8 x = - x ⟹ x^{2} - 7 x = 0 ⟹ x = 0$ or $x = 7$ . The area is $\int_{0}^{7} (7 x - x^{2}) d x = {[\frac{7 x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{7} = \frac{7 (49)}{2} - \frac{343}{3} = 343 (\frac{1}{2} - \frac{1}{3}) = \frac{343}{6}$ .
Verify: Archimedes formula for area enclosed by a parabola and a chord: Area $= \frac{1}{6} ∣ a ∣ (x_{2} - x_{1})^{3} = \frac{1}{6} (1) (7 - 0)^{3} = \frac{343}{6}$ , which matches.

Shortcut:

When to use it: Whenever finding the area enclosed between a vertical parabola $y = a x^{2} + b x + c$ and a secant line

Given: Parabola $y = x^{2} - 8 x$ and line $y = - x$ , with coefficient of $x^{2}$ being $a = 1$ .

Goal: Compute area using Archimedes formula Area $= \frac{∣ a ∣}{6} (Δ x)^{3}$ .

Approach: Find difference of roots $Δ x$ from equation $x^{2} - 7 x = 0$ , which gives $Δ x = 7 - 0 = 7$ .

Execute: Area $= \frac{1}{6} \times 1 \times 7^{3} = \frac{343}{6}$ .

Verify: Denominator is 6 and numerator is $7^{3} = 343$ , directly matching Option A.

How do you find the area bounded by a parabola and a line?

What feels right?

The enclosed region is bounded between $x = 0$ and $x = 7$ , yielding an area of $\frac{343}{6}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

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