Let r be the region bounded by the graphs of #y= sqrt(x)# and #y=x/2#, how do you find the area of R?

Question

Emily Ammerman · Accepted Answer

The area of R #=10/3# Let's find the intercepts of the two graphs. #sqrtx=x/2##=>##x=x^2/4# #=>##x(x/4-1)# so, #x=0# and #x=4# a small area is #dA=(sqrtx-x/4)dx# #:. A=int_0^(4)(sqrtx-x/4)dx# #=[2/3x^(3/2)-x^2/8]_0^4=16/3-2=10/3 # graph{(y-sqrtx)(y-x/2)=0 [-4.19, 6.907, -2.38, 3.167]}

Elijah Abram · Answer

undefined To find the area of the region $R$ bounded by the graphs of $y = \sqrt{x}$ and $y = \frac{x}{2}$, you need to first determine the points of intersection of the two graphs.

Set $ \sqrt{x} = \frac{x}{2} $ and solve for $x$. This gives $ x = 4 $. So, the points of intersection are at $ x = 0 $ and $ x = 4 $.

The area $ A $ of $ R $ is then given by the integral of the difference between the two functions from $x = 0$ to $x = 4$:

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

This simplifies to:

$$ A = \int_{0}^{4} \left( \frac{x}{2} - x^{1/2} \right) dx $$

$$ A = \int_{0}^{4} \left( \frac{1}{2}x - x^{1/2} \right) dx $$

$$ A = \left[ \frac{1}{4}x^2 - \frac{2}{3}x^{3/2} \right]_{0}^{4} $$

$$ A = \left( \frac{1}{4}(4)^2 - \frac{2}{3}(4)^{3/2} \right) - \left( \frac{1}{4}(0)^2 - \frac{2}{3}(0)^{3/2} \right) $$

$$ A = \left( \frac{1}{4}(16) - \frac{2}{3}(8) \right) - \left( \frac{1}{4}(0) - \frac{2}{3}(0) \right) $$

$$ A = \left( 4 - \frac{16}{3} \right) - 0 $$

$$ A = \frac{12}{3} - \frac{16}{3} $$

$$ A = -\frac{4}{3} $$

So, the area of the region $R$ is $ \frac{4}{3} $.