Area between cos(x) and -x^3 + 4x^2

[fdragon]

Two functions that intersect 3 times are f(x)= cos(x) and g(x) = -x³ + 4x². Find the area between the curves.

so I got.....

y = cos(x)
y/cos = x

and

y = -x³ + 4x²
y^(1/3) = -x + 4x²
(y^(1/3))^(1/2) = -x + 4x
((y^(1/3))^(1/2))/4 = -x + x
-((y^(1/3))^(1/2))/4 = x+x

how do I finish this???

and for the bounds... do I just look at the graph or is there a way to calculate it?

Then I would take the Integral (w/bounds) of (Right fuction - Left Function) dy.... right?

 

[stapel]

y = cos(x)
y/cos = x

From what you've posted, it would appear that you are in a calculus class that depends upon trigonometry, but you've never taken trig, because you appear to be under the impression that "cos(x)" means "cos" multiplied by "x", rather than "f(x) = cos(x)". (That is to say, "cos()" represents a named function.)

Since your calculus course requires trig, you really need to take a trig class before taking the calculus class. It might be wise to conference with your academic advisor about appropriate placement.

My best wishes to you.

Eliz.

 

[skeeter]

this area problem was meant to be done using technology ... it would be extremely difficult to determine the x-values where cos(x) intersects 4x² - x³ by hand.

so ... get out your graphing calculator, graph the two functions, find the x-values for their 3 points of intersection (and store them). Then, set up an expression consisting of one or more integrals that will yield the desired area.

 

[fdragon]

Ok I got it... I was using the wrong method..... I did it another way and it worked.