I'm sorry maybe it's obvious but English is not my first language. I just want to know what is asked in this question:
The area of a circle (in square inches) is numerically larger than its circumference (in inches). What is the smallest possible integral area of the circle, in square inches?
Specifically I don't understand what integral area means. I'm familiar with integration and how you can calculate the area between the curve and the axes but what should the integral area of a circle mean?
On
As Wouter points out, "integral" almost certainly means "integer valued". That is, the area of the circle is an integer. We can now answer the question, which I likely would have written as
The area of a circle (in square inches) is numerically larger than its circumference (in inches). What is the smallest integer which could be the area of the circle (in square inches)?
To fix notation, suppose that we have a circle with area and circumference given by
$$ \text{Area} = A \text{ in}^2 \qquad\text{and}\qquad \text{Circumference} = C \text{ in}. $$
The first sentence tells us that $A > C$. The second sentence asks us to find the smallest integer value of $A$ possible. From general theory, we know that if $r$ inches is the radius of the circle, then $$ A = \pi r^2 \qquad\text{and}\qquad C = 2\pi r. $$ This implies that $$ A = \frac{C^2}{4\pi}. $$ Since we need $A > C$ (and we can assume that $C > 0$), it follows that $$ \frac{C^2}{4\pi} > C \implies C > 4\pi \implies A = \frac{C^2}{4\pi} > \frac{(4\pi)^2}{4\pi} = 4\pi, $$ since $C > 4\pi > 1$ implies that $C^2 > (4\pi)^2$. But then (1) $A$ has to be an integer and (2) $A$ must be bigger than $4\pi$, so we round up to obtain $$ A = \lceil 4\pi \rceil = 13 $$ (since $4\pi \approx 12.566$; thanks Google!). That is, the smallest integer which could be the area of the circle is 13 square inches.
Most likely, "integral" means "is an integer". The area of this circle, expressed in square inches, is an integer.