Integral Calculus – Area Under Curves – JEE Main 28 Jan 2026 Shift 1

Solution:

We need to find the area bounded by:
1. $xy = 8 \Rightarrow y = 8/x$
2. $y = x^2$
3. $y = 1$

First, find the intersection points of the curves.
Intersection of $y = x^2$ and $y = 1$:
$$x^2 = 1 \Rightarrow x = 1 \quad (\text{since } x \ge 0)$$
Intersection of $y = x^2$ and $xy = 8$:
$$x(x^2) = 8 \Rightarrow x^3 = 8 \Rightarrow x = 2$$
Intersection of $xy = 8$ and $y = 1$:
$$x(1) = 8 \Rightarrow x = 8$$


The region is split into two parts based on the upper boundary:
1. From $x = 1$ to $x = 2$, the upper curve is $y = x^2$ and lower is $y = 1$.
2. From $x = 2$ to $x = 8$, the upper curve is $y = 8/x$ and lower is $y = 1$.

Area $A = \int_{1}^{2} (x^2 – 1) dx + \int_{2}^{8} (\frac{8}{x} – 1) dx$.

Calculate the first integral:
$$\left[ \frac{x^3}{3} – x \right]_{1}^{2} = \left(\frac{8}{3} – 2\right) – \left(\frac{1}{3} – 1\right)$$
$$= \frac{2}{3} – \left(-\frac{2}{3}\right) = \frac{4}{3}$$

Calculate the second integral:
$$\left[ 8 \ln x – x \right]_{2}^{8} = (8 \ln 8 – 8) – (8 \ln 2 – 2)$$
$$= 8(3 \ln 2) – 8 – 8 \ln 2 + 2$$
$$= 24 \ln 2 – 8 \ln 2 – 6$$
$$= 16 \ln 2 – 6$$

Total Area:
$$A = \frac{4}{3} + 16 \ln 2 – 6$$
$$A = 16 \ln 2 – \frac{14}{3}$$
$$A = \frac{48 \ln 2 – 14}{3} = \frac{2}{3}(24 \ln 2 – 7)$$

Ans. (3)