JEE MainMathematics

Answer verified against the official NTA key

+4 marks−1 if incorrectSingle correctPrevious-year question

Area of Bounded Region Between a Hyperbola and a Parabola

The area of the region given by {(x,y): xy ≤ 8, 1 ≤ y ≤ x^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.

A16log_e 2+7/3B16log_e 2-14/3C8log_e 2-13/3D8log_e 2+7/6

Step-by-step solutionView

Correct answer

The bounded area is found by integrating $(x^{2} - 1)$ from $x = 1$ to $x = 2$ and $(\frac{8}{x} - 1)$ from $x = 2$ to $x = 8$ , yielding $16 \log_{e} 2 - \frac{14}{3}$ .

Option analysis

Why each option works or fails

A · 16log_e 2+7/3

Adding the lower boundary integral instead of subtracting it, or miscalculating the polynomial component as positive $\frac{7}{3}$ . Subtract the lower boundary $y = 1$ from the upper boundary functions across the entire interval of integration.

B · 16log_e 2-14/3

This is the correct calculation: $\int_{1}^{2} (x^{2} - 1) d x + \int_{2}^{8} (\frac{8}{x} - 1) d x = \frac{4}{3} + (16 \log_{e} 2 - 6) = 16 \log_{e} 2 - \frac{14}{3}$ . Ensure both sub-intervals are integrated correctly with respect to $x$ (or integrate with respect to $y$ ).

C · 8log_e 2-13/3

Integrating with an upper limit of $x = 4$ instead of $x = 8$ for the hyperbola, leading to $8 \log_{e} 2$ instead of $16 \log_{e} 2$ . Find the correct intersection of the hyperbola $y = \frac{8}{x}$ and the line $y = 1$ , which gives $x = 8$ .

D · 8log_e 2+7/6

Evaluating the logarithmic term as $8 \log_{e} (8 / 4)$ or dropping the factor of $8$ while incorrectly combining the constant terms. Carefully compute $\int_{2}^{8} \frac{8}{x} d x = 8 [\ln x]_{2}^{8} = 8 \ln (4) = 16 \ln 2$ .

Reviewed route

Solution

StepWorking

01Given

The region is defined by $x y \leq 8$ , $y \geq 1$ , and $y \leq x^{2}$ with $x, y > 0$ .

02Goal

Find the area of the enclosed region $A$ .

03Approach

Find the intersection points of the bounding curves $y = x^{2}$ (or $x = \sqrt{y}$ ) and $x y = 8$ (or $x = 8 / y$ ). For $y \geq 1$ , the right boundary is $x = 8 / y$ and the left boundary is $x = \sqrt{y}$ . Integrate with respect to $y$ from $y = 1$ to the intersection point $y = 4$ .

04Execute

Solve $x^{2} \cdot x = 8 ⟹ x^{3} = 8 ⟹ x = 2$ , which gives $y = 4$ . The integral is $A = \int_{1}^{4} (\frac{8}{y} - \sqrt{y}) d y = {[8 \ln y - \frac{2}{3} y^{3 / 2}]}_{1}^{4} = 8 \ln 4 - \frac{2}{3} (4^{3 / 2} - 1) = 16 \ln 2 - \frac{14}{3}$ .

✓Verify

Check limits: at $y = 1$ , $\sqrt{y} = 1$ and $8 / y = 8$ , so width is $7$ . At $y = 4$ , both curves meet at $x = 2$ . The integrand $\frac{8}{y} - \sqrt{y} > 0$ on $(1, 4)$ , confirming positive area $16 \ln 2 - 14 / 3 \approx 11.09 - 4.67 = 6.42 > 0$ .

Hints that build this answer step by step

What are the boundary curves and intersection points that delimit the region in the first quadrant?
The region is bounded below by $y = 1$ from $x = 1$ to $x = 8$ , and bounded above by $y = x^{2}$ on $[1, 2]$ and $y = \frac{8}{x}$ on $[2, 8]$ .
Which setup correctly splits the region into vertical slices to integrate with respect to $x$ ?
$\int_{1}^{2} (x^{2} - 1) d x + \int_{2}^{8} (\frac{8}{x} - 1) d x$
What is the value of $\int_{1}^{2} (x^{2} - 1) d x + \int_{2}^{8} (\frac{8}{x} - 1) d x$ ?
$16 \log_{e} 2 - \frac{14}{3}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 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 ( $x \in [1, 2]$ and $x \in [2, 8]$ ), whereas integrating along y gives a single integral from $y = 1$ to $y = 4$ .

Answer

The bounded area is found by integrating $(x^{2} - 1)$ from $x = 1$ to $x = 2$ and $(\frac{8}{x} - 1)$ from $x = 2$ to $x = 8$ , yielding $16 \log_{e} 2 - \frac{14}{3}$ .

Why each option works or fails

A: 16log_e 2+7/3 - Adding the lower boundary integral instead of subtracting it, or miscalculating the polynomial component as positive $\frac{7}{3}$ . Subtract the lower boundary $y = 1$ from the upper boundary functions across the entire interval of integration.
B · correct: 16log_e 2-14/3 - This is the correct calculation: $\int_{1}^{2} (x^{2} - 1) d x + \int_{2}^{8} (\frac{8}{x} - 1) d x = \frac{4}{3} + (16 \log_{e} 2 - 6) = 16 \log_{e} 2 - \frac{14}{3}$ . Ensure both sub-intervals are integrated correctly with respect to $x$ (or integrate with respect to $y$ ).
C: 8log_e 2-13/3 - Integrating with an upper limit of $x = 4$ instead of $x = 8$ for the hyperbola, leading to $8 \log_{e} 2$ instead of $16 \log_{e} 2$ . Find the correct intersection of the hyperbola $y = \frac{8}{x}$ and the line $y = 1$ , which gives $x = 8$ .
D: 8log_e 2+7/6 - Evaluating the logarithmic term as $8 \log_{e} (8 / 4)$ or dropping the factor of $8$ while incorrectly combining the constant terms. Carefully compute $\int_{2}^{8} \frac{8}{x} d x = 8 [\ln x]_{2}^{8} = 8 \ln (4) = 16 \ln 2$ .

Step-by-step solution

Given: The region is defined by $x y \leq 8$ , $y \geq 1$ , and $y \leq x^{2}$ with $x, y > 0$ .
Goal: Find the area of the enclosed region $A$ .
Approach: Find the intersection points of the bounding curves $y = x^{2}$ (or $x = \sqrt{y}$ ) and $x y = 8$ (or $x = 8 / y$ ). For $y \geq 1$ , the right boundary is $x = 8 / y$ and the left boundary is $y = \frac{8}{x}$ . Integrate with respect to $y$ from $y = 1$ to the intersection point $y = 4$ .
Execute: Solve $8 \log_{e} (8 / 4)$ , which gives $y = 4$ . The integral is $A = \int_{1}^{4} (\frac{8}{y} - \sqrt{y}) d y = {[8 \ln y - \frac{2}{3} y^{3 / 2}]}_{1}^{4} = 8 \ln 4 - \frac{2}{3} (4^{3 / 2} - 1) = 16 \ln 2 - \frac{14}{3}$ .
Verify: Check limits: at $y = 1$ , $\sqrt{y} = 1$ and $8 / y = 8$ , so width is $7$ . At $y = 4$ , both curves meet at $x = 2$ . The integrand $\frac{8}{y} - \sqrt{y} > 0$ on $(1, 4)$ , confirming positive area $16 \ln 2 - 14 / 3 \approx 11.09 - 4.67 = 6.42 > 0$ .

Area of Bounded Region Between a Hyperbola and a Parabola

What feels right?

The bounded area is found by integrating $(x^{2} - 1)$ from $x = 1$ to $x = 2$ and $(\frac{8}{x} - 1)$ from $x = 2$ to $x = 8$ , yielding $16 \log_{e} 2 - \frac{14}{3}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

We think you should solve this next.

The same chapter, from another paper — Integral Calculus, 2023

The same chapter, from another paper — Integral Calculus, 2023

The point of intersection $C$ of the plane $8 x + y + 2 z = 0$ and the line joining th…

Students also ask

Area of Bounded Region Between a Hyperbola and a Parabola

The bounded area is found by integrating (x2−1)(x2−1) from x=1x=1 to x=2x=2 and (8x−1)(x8​−1) from x=2x=2 to x=8x=8, yielding 16log⁡e2−14316loge​2−314​.

01Given

02Goal

03Approach

04Execute

✓Verify

The same chapter, from another paper — Integral Calculus, 2023

The same chapter, from another paper — Integral Calculus, 2023

The point of intersection CC of the plane 8x+y+2z=08x+y+2z=0 and the line joining th…

Students also ask

The bounded area is found by integrating $(x^{2} - 1)$ from $x = 1$ to $x = 2$ and $(\frac{8}{x} - 1)$ from $x = 2$ to $x = 8$ , yielding $16 \log_{e} 2 - \frac{14}{3}$ .

The point of intersection $C$ of the plane $8 x + y + 2 z = 0$ and the line joining th…