Area bounded by curve: wrong answer?

[cyt91]

Find the area of region bounded by y = 4 -x^2 ; y = 2 - x; x = - 2 and x = 3

I've calculated the area to be 11/3 however the answer given is 49/6.
Is the answer correct? I've recalculated this twice and there seems to be nothing wrong with my working.

https://skydrive.live.com/?cid=6b041751c72e14ad#!/?cid=6b041751c72e14ad&sc=photos&uc=3&id=6B041751C72E14AD%21149!cid=6B041751C72E14AD&id=6B041751C72E14AD%21153&sc=photos

https://skydrive.live.com/?cid=6b041751c72e14ad#!/?cid=6b041751c72e14ad&sc=photos&uc=3&id=6B041751C72E14AD%21149!cid=6B041751C72E14AD&id=6B041751C72E14AD%21151&sc=photos

https://skydrive.live.com/?cid=6b041751c72e14ad#!/?cid=6b041751c72e14ad&sc=photos&uc=3&id=6B041751C72E14AD%21149!cid=6B041751C72E14AD&id=6B041751C72E14AD%21152&sc=photos

Thank you.

 

[Tomer]

Did you check where the functions cut each other? Notice that the intersection points are different than the integrals boundaries, which means you'd have to do several different calculations...

 

[tiny-tim]

hi cyt91!

Find the area of region bounded by y = 4 -x^2 ; y = 2 - x; x = - 2 and x = 3.

that question doesn't look right …

can you check it?​

 

[cyt91]

Yea,I did.
The intersection points occur at x=2 and x=-1.Please view the images I've posted in the first post. They are my workings.The question is as given in my worksheet. I did not alter it.

Thanks.

 

[Tomer]

Sorry, I cannot see the pictures you've uploaded.

I'm getting 43/6, lol :-)

Edit: ok, I've managed to see your pics.
I have two questions:
1. Why are you ignoring the the middle integral, from -1 to 2? In this part the parabola is above the line, and so you need to calculate this part as well.
2. Why did you put an absolute value on the last integral? The absolute value in the end is unnecessary. The area between two functions, assuming f(x)>=g(x) in that interval, is always int(f(x)-g(x)). It'll always be positive.
It is of course not that important in this question, but it may be important on others, where one of the integral limits could be a variable, for example...

I'm too lazy to check myself, but what I got is 43/6

 

[cyt91]

Ok. So we cannot ignore the middle part bounded by the parabola and straight line...
I thought we must only include the area bounded by parabola,straight line and (x=-3 or x=2). Coz the question states we must find the area of the regions bounded by parabola,straight line,x=-3 and x=2. so...

I took the absolute value because area under the x-axis would be negative.(that's what I learned in high school.) Apparently that's not the case here.

Thank you for answering my question.

 

[Tomer]

Ok. So we cannot ignore the middle part bounded by the parabola and straight line...
I thought we must only include the area bounded by parabola,straight line and (x=-3 or x=2). Coz the question states we must find the area of the regions bounded by parabola,straight line,x=-3 and x=2. so...

I took the absolute value because area under the x-axis would be negative.(that's what I learned in high school.) Apparently that's not the case here.

Thank you for answering my question.

Well, usually when one says: find the area between the functions f(x) and g(x) in the interval [a,b], one means the whole area trapped between the functions in this interval.
I guess one could interpret the question the way you have (it might even be more accurate). I guess you just need to see which calculation fits their answer. But as I've said, the convention is not ignoring the middle part...

No prob :-)