Sketch the region enclosed by $y= \sqrt{x}$, $y=0$, $x=4$ and calculate its area

Question

I do not understand how this would work to find the area. Any help would be appreciated.

Answer 1

You can sketch the graph on paper. The required area is area under curve of $\sqrt{x}$ from $x=0$ to $x=4$.

$$\int_{0}^{4} \sqrt{x} dx = \frac{2}{3} \left(4^{3/2}-0^{3/2}\right) = \frac{16}{3}$$

Answer 2

Hint:

Won't that simply be $$\int _{0}^4 \sqrt x. dx=?$$

Answer 3

I am sure you can sketch the graph.

Note that there are two different ways to find the area.

One way which is probably easier, is to avoid the square root and integrate in y direction.

Area =$ \int\limits_0^2(4-y^2)dy$= $16/3$

The other way is to integrate in x direction and remember to change your limits of integration to $0$ and $4$