The nearest integral value of the shaded area shown below is:

The shaded area $A$ is defined by the integral of the function $y = \sqrt{x}$ from $x = 1$ to $x = 4$.

$A = \int_{1}^{4} \sqrt{x} \, dx$

---

The integral of $x^{1/2}$ is calculated using the power rule:

$\int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}$

---

Evaluating the expression at the upper and lower limits:

$A = [\frac{2}{3}x^{3/2}]_{1}^{4}$

$A = \frac{2}{3}(4)^{3/2} - \frac{2}{3}(1)^{3/2}$

$A = \frac{2}{3}(8) - \frac{2}{3}(1)$

$A = \frac{16}{3} - \frac{2}{3}$

$A = \frac{14}{3}$

The area is $\frac{14}{3}$ square units.

---

Note: Yes, the question said "nearest integral value", the answer should have been 6 square units. However, it is NTA's world and we're just living in it.

Here's proof (please don't take it out on us, we're as frustrated as you are):