Finding the Area Between Curves

Are you sure that is all the information given? I'm not finding an area between the curves. Do you mean between them from 0 to 4?

In this case you need to subtract the lower function from the higher one and THEN integrate from 0 to 4.

Hope this helps.
 
No, it is not from 0 to 4 I just did that after I got the integral. The only info that was given was the Y values and it said to find the area between the curves. I will edit my post so there is no confusion.
 
CatchThis2 said:
Find the area between the curves of y=x^3-11x^2+28x , y=-x^3+11x^2-28x
Click to expand...

So, set the two equations equal to each other in order to find the x values where they intersect. Then do what I described in my last post.
 
\(\displaystyle A \ = \ \int_{0}^{4}[2x^{3}-22x^{2}+56x]dx \ + \ \int_{4}^{7}[-2x^{3}+22x^2-56x]dx \ = \ \frac{937}{6}sq. \ units\)
 
BigGlenntheHeavy said:
\(\displaystyle A \ = \ \int_{0}^{4}[2x^{3}-22x^{2}+56x]dx \ + \ \int_{4}^{7}[-2x^{3}+22x^2-56x]dx \ = \ \frac{937}{6}sq. \ units\)
Click to expand...

This is why people expect the answers :evil:
 
\(\displaystyle JuicyBurger:\)

\(\displaystyle It \ is \ setting \ up \ the \ problem \ that \ requires \ skill, \ as \ the \ answer \ then \ is \ academic.\)
 
How did you get the derivative of x^3-11x^2+28x. Maybe someone can show me step by step so I can follow. I really appreciate the feedback. I am very impressed with this site and its members as you all have been so nice and helpful to a new member like myself. I appreciate everything everyone has done for me!

Thanks,

GARRETT
 
\(\displaystyle Hey, GARRETT, \ what \ you \ see \ is \ what \ you \ get.\)

\(\displaystyle You \ figure \ it \ out, \ a \ good \ exercise \ for \ you.\)
 
JuicyBurger is right! But for better understanding.
The info given is the problem is sufficient.
To find where the equations meet which will be the limits of integ we need to equate the the equations and solve them. Since one is exact negative of other you will get a mirror image
x^3-11x^2+28x=-(x^3-11x^2+28x)
Solving this we get x=0,4,7
So this means that the curve pass through x axis at these points and intersect each other.
And now to find area between the curves we take
Area=2*[{integ(x^3-11x^2+28x) from 0 to 4} + {integ(-x^3+11x^2-28x) from 4 to 7}]
We multiply it with 2 because the integration is between the curve and x-axis. And between 0 and 4, (x^3-11x^2+28x) is above x-axis and from 4 to 7, (-x^3+11x^2-28x).
Hope this helps! :)
I am sorry I couldnt get the expressions in a more presentable form
 
BigGlenntheHeavy said:
\(\displaystyle JuicyBurger:\)

\(\displaystyle It \ is \ setting \ up \ the \ problem \ that \ requires \ skill, \ as \ the \ answer \ then \ is \ academic.\)
Click to expand...

I think that it is obvious that Juicy Burger was not thinking 937/6 when writing "expect the answers".

If you claim that setting up the integrals requires the greater skill, big guy, then why aren't you helping to guide the original poster through that process versus spoon-feeding?
 
BigGlenntheHeavy said:
\(\displaystyle Hey, GARRETT, \ what \ you \ see \ is \ what \ you \ get.\)

\(\displaystyle You \ figure \ it \ out, \ a \ good \ exercise \ for \ you.\)
Click to expand...

Heh, heh.

The original poster just told you, big guy, that he thinks that you provided him with the derivative of a cubic polynomial.

In light of that, your response above is worthless.

And, what's up with your new signature? Did chrisr sting you so deeply that you simply cannot let it go?
 
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