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

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The area of region R, bounded by the functions f(x)=3√x−4 and g(x)=3x/5−8/5, can be determined by first finding the roots of the equation 3√x−4=3x/5−8/5. These roots serve as the endpoints for the integral used to calculate the area A of R. The correct formulation for the area is given by the integral from a to b of the difference between the two functions. Clarifications were made regarding the definitions of the functions involved, ensuring accurate calculations. The discussion emphasizes the importance of correctly identifying the functions to achieve the desired area calculation.
sgalos05
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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.
 
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Beer soaked query follows.
sgalos05 said:
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?
 
sgalos05 said:
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 $R$.
 
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 $g(x)= \frac{3x}{5}- \frac{8}{5}= \frac{3x- 8}{5}$.
 
Just spitballin’ here ...

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