# 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?

The area of R

By signing up, you agree to our Terms of Service and Privacy Policy

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} ).

By signing up, you agree to our Terms of Service and Privacy Policy

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7