Set up integral for area bounded by the graphs of---
y = 13x^3 - 5x^2 + 6x - e^cos(10x)
y = 4x^3 + x^2 + 14x - e^cos(10x)
Now... I feel as though I'm almost done.... and yet in the problem it was stated we can stop working on it when we have eliminated absolute values.... I have yet to even see a need for absolute values.... isn't the Area for the two functions, the integral of f(x) - g(x)? doesn't that eliminate the e^cos(10x) parts?

Question

Set up integral for area bounded by the graphs of---
y = 13x^3 - 5x^2 + 6x - e^cos(10x)
y = 4x^3 + x^2 + 14x - e^cos(10x)
Now... I feel as though I'm almost done.... and yet in the problem it was stated we can stop working on it when we have eliminated absolute values.... I have yet to even see a need for absolute values.... isn't the Area for the two functions, the integral of f(x) - g(x)? doesn't that eliminate the e^cos(10x) parts?

anonymous · Answer

yes that term will be eliminated ..

anonymous · Answer

|dw:1347603783233:dw|

anonymous · Answer

Area for the two functions, the integral of f(x) - g(x)?
here it's fine..|dw:1347603955932:dw|

anonymous · Answer

so rough sketch is needed for finding the area enclosed by two curves..

anonymous · Answer

but in ur problem..Area for the two functions, the integral of f(x) - g(x)...will work

anonymous · Answer

our problem said a graph was not needed, so I assumed it was a basic one where it's the integral of (f(x)) - g(x)... so I found the intercepts to determine the {a,b} part of the integral, took the antiderivative.... and found it to be a basic function, so I'm a little worried when it says stop when you get rid of absolute values.... since you'd think our professo put it there for a reason...

anonymous · Answer

you get rid of absolute values???

anonymous · Answer

.... I don't know. Apparently we're supposed to come to a point where there's absolute values.....and I'm not seeing where that would be or why.....

anonymous · Answer

have u got the final  numeric value of area enclosed by graphs?

anonymous · Answer

I mean ...integral of f(x) - g(x)..and then putting the limits.[a,b]

anonymous · Answer

absolute value...I think...we are just doing integral of f(x) - g(x) without plotting the graphs..might be...we can get negative area...but area cant be take...sow e take absolute value and that's the final answer.

anonymous · Answer

|dw:1347604781755:dw|

anonymous · Answer

Yeah, the portion intersects in a negative quadrant.... so would I put said absolute values in the integral, or do they go into the equation after I take the antiderivative?

anonymous · Answer

Well, I must have done something wrong, because it came out equal tozero.... and being that there IS an area sectioned off by those two functions, pretty sure the area is not zero...

anonymous · Answer

u have to take the absolute value after evaluating the definite integral..$$\left|  \int\limits_{a}^{b}  f(x) \right| $$

anonymous · Answer

can u tell me the value of a and b...where the curves are intersecting