MHB Determine the area of a region between two curves defined by algebraic functions

[sgalos05]

R is the region bounded by the functions f(x)=3√x−4 and g(x)=3x/5−8/5. Find the area A of R. Enter answer using exact values.

 

[jonah1]

Beer soaked query follows.

R is the region bounded by the functions f(x)=3√x−4 and g(x)=3x/5−8/5. Find the area A of R. Enter answer using exact values.

What have you done so far?

 

[Greg]

R is the region bounded by the functions f(x)=3√x−4 and g(x)=3x/5−8/5. Find the area A of R. Enter answer using exact values.

You need the roots $a,b$ of the equation $3\sqrt{x}-4=3x\sqrt{5}-\frac{8}{5}$

Once you have established these roots use them as endpoints in

$\int_{a}^{b}\left(3\sqrt{x}-4-3x\sqrt{5}+\frac{8}{5}\right)dx$

The result is the area $A$ of $R$.

 

[HOI]

Is the first first function $f(x)= \sqrt{x}- 4$ or $f(x)= \sqrt{x- 4}$?

Also I do not see the second function as $g(x)= 3x\sqrt{5}- \frac{8}{5}$. I see $g(x)= \frac{3x}{5}- \frac{8}{5}= \frac{3x- 8}{5}$.

 

[skeeter]

Just spitballin’ here ...

I would say $f(x) = 3\sqrt{x} - 4$ since it yields a simpler, nice solution when equating it to $g(x)$

 

[Greg]

Is the first first function $f(x)= \sqrt{x}- 4$ or $f(x)= \sqrt{x- 4}$?

Also I do not see the second function as $g(x)= 3x\sqrt{5}- \frac{8}{5}$. I see $g(x)= \frac{3x}{5}- \frac{8}{5}= \frac{3x- 8}{5}$.

yeah should have zoomed the page... lol... :)