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Integral Calculus: JEE Main Mathematics Question with Solution

The area of the region given by

\left\{(x,y): xy \le 8, 1 \le y \le x^2\right\}

is :

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16\log_e 2+\frac{7}{3}

16\log_e 2-\frac{14}{3}

8\log_e 2-\frac{13}{3}

8\log_e 2+\frac{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 $\left(\frac{8}{x} - 1\right)$ from $x=2$ to $x=8$ , yielding $16\log_e 2 - \frac{14}{3}$ .

Option analysis

Why each option works or fails

A ·

16\log_e 2+\frac{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 ·

16\log_e 2-\frac{14}{3}

None. This is the correct calculation: $\int_1^2 (x^2 - 1)\,dx + \int_2^8 \left(\frac{8}{x} - 1\right)dx = \frac{4}{3} + (16\log_e 2 - 6) = 16\log_e 2 - \frac{14}{3}$ . Ensure both sub-intervals are integrated correctly with respect to $x$ (or integrate with respect to $y$ ).

C ·

8\log_e 2-\frac{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 ·

8\log_e 2+\frac{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 $\int_2^8 \frac{8}{x}\,dx = 8[\ln x]_2^8 = 8\ln(4) = 16\ln 2$ .

Reviewed route

Solution

StepWorking

01given

The region is defined by $xy \le 8$ , $y \ge 1$ , and $y \le x^2$ 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 $xy = 8$ (or $x = 8/y$ ). For $y \ge 1$ , 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 $x^2 \cdot x = 8 \implies x^3 = 8 \implies x = 2$ , which gives $y = 4$ . The integral is $A = \int_1^4 \left(\frac{8}{y} - \sqrt{y}\right) dy = \left[ 8 \ln y - \frac{2}{3} y^{3/2} \right]_1^4 = 8 \ln 4 - \frac{2}{3}(4^{3/2} - 1) = 16 \ln 2 - \frac{14}{3}$ .

✓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 $16\ln 2 - 14/3 \approx 11.09 - 4.67 = 6.42 > 0$ .

✓ 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 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 $\left(\frac{8}{x} - 1\right)$ from $x=2$ to $x=8$ , yielding $16\log_e 2 - \frac{14}{3}$ .

Why each option works or fails

A: $16\log_e 2+\frac{7}{3}$ - 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: $16\log_e 2-\frac{14}{3}$ - None. This is the correct calculation: $\int_1^2 (x^2 - 1)\,dx + \int_2^8 \left(\frac{8}{x} - 1\right)dx = \frac{4}{3} + (16\log_e 2 - 6) = 16\log_e 2 - \frac{14}{3}$ . Ensure both sub-intervals are integrated correctly with respect to $x$ (or integrate with respect to $y$ ).
C: $8\log_e 2-\frac{13}{3}$ - 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: $8\log_e 2+\frac{7}{6}$ - Evaluating the logarithmic term as $8\log_e(8/4)$ or dropping the factor of $8$ while incorrectly combining the constant terms. Carefully compute $\int_2^8 \frac{8}{x}\,dx = 8[\ln x]_2^8 = 8\ln(4) = 16\ln 2$ .

Step-by-step solution

given: The region is defined by $xy \le 8$ , $y \ge 1$ , and $y \le x^2$ 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 $xy = 8$ (or $x = 8/y$ ). For $y \ge 1$ , 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 $y = 4$ .
execute: Solve $x^2 \cdot x = 8 \implies x^3 = 8 \implies x = 2$ , which gives $y = 4$ . The integral is $A = \int_1^4 \left(\frac{8}{y} - \sqrt{y}\right) dy = \left[ 8 \ln y - \frac{2}{3} y^{3/2} \right]_1^4 = 8 \ln 4 - \frac{2}{3}(4^{3/2} - 1) = 16 \ln 2 - \frac{14}{3}$ .
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 $16\ln 2 - 14/3 \approx 11.09 - 4.67 = 6.42 > 0$ .

Integral Calculus: JEE Main Mathematics Question with Solution

The bounded area is found by integrating (x2−1)(x^2 - 1)(x2−1) from x=1x=1x=1 to x=2x=2x=2 and (8x−1)\left(\frac{8}{x} - 1\right)(x8​−1) from x=2x=2x=2 to x=8x=8x=8, yielding 16log⁡e2−14316\log_e 2 - \frac{14}{3}16loge​2−314​.

01given

02goal

03approach

04execute

✓verify

Students also ask

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