Integration - Area Question

Dr.Millah

Find the points of intersection of the curves y = x^2 +1 and y = 7 − x .
Find the area enclosed by the two curves.

Not fully sure about this question, usually they seem to state between x=1 - x=4 or something similar. Any help would be appreciated.

Thanks

blubloblu

Sketch it

bnt

The "between" points will be the intersection points - you have to work those out first.

Dr.Millah

bnt wrote: »

The "between" points will be the intersection points - you have to work those out first.

so solve x and y = 0?

Hammer Archer

To find the points of intersection, let your two equations above equal each other and solve for 2 values of x. These will be your "between" points.
I'd also suggest drawing a rough sketch of it as it'll be much easier to visualise what the enclosed area looks like.

bnt

Then when you have them, you work out the definite integral of each original equation between those points and subtract the one from the other.

The sketch should tell you which to subtract from which, but if you get that wrong, all that happens is you get a minus sign in the result. If this is for an assignment, you might get marked on that, so I think it's best draw the sketch to find out which formula marks the upper boundary of the area.

Dr.Millah

So i equaled the equations to each other and got two x values
x= -3
x= 1

so the intersection points are (-3,0) and (2,0) yea?

Or, do i just intergrate the two equations i was given and use 2 and -3 as the upper and lower limits and add the two areas?

LeixlipRed

How did you get those values? They're wrong yet close to the actual answer. The issue here is that you don't seem to understand the relationship between area under a curve and integration. Read up on that then sketch the two curves. Do you know how to do that?

Dr.Millah

LeixlipRed wrote: »

How did you get those values? They're wrong yet close to the actual answer. The issue here is that you don't seem to understand the relationship between area under a curve and integration. Read up on that then sketch the two curves. Do you know how to do that?

Just corrected myself there. Made a silly mistake in the formula (-b+-....)

I know how to do the area problems that involve one curve and you are given the limits, however i am not fully sure how to work this problem out. Normally when i graph the curve it will say between x=-1 to x=2 etc. Not sure what i do with this question

so, intersection points are (2,5)(-3,10)

LeixlipRed

Which means you don't fully understand it. Have you tried to sketch the curves?

Dr.Millah

LeixlipRed wrote: »

Which means you don't fully understand it. Have you tried to sketch the curves?

yea,
for the curve i have points y = x^2+1
(-2,5)
(-1,2)
(0,1)
(1,2)
(2,5)
(3,10)


and the line y = 7-x

(1,6)
(2,5)
(3,4) etc

Not sure what i should be doing though

LeixlipRed

Yes, but have you tried drawing it on a piece of paper? Then read the original question, decide what area it is you have to find and then see how that relates to the points of intersection of the two curves.

Dr.Millah

LeixlipRed wrote: »

Yes, but have you tried drawing it on a piece of paper? Then read the original question, decide what area it is you have to find and then see how that relates to the points of intersection of the two curves.

Yes, on paper it is clear what area i am looking for. Not fully sure what way to find it.

Bit of tinkering around.. Integrated the line formula and work it out using the upper limit 2 and lower -3 for an answer of 37.5

Integrated the curve with same limits and got answer 16.66.

37.5-16.66 = 20.84 area of shaded region

I think i have it now. But i could be wrong

Fremen

I haven't checked your numerics but your idea is correct, good job.
Here's an accurate plot of the curves you were looking at. Might help to build yourt intuition. From your sketch, it looks like you missed the fact that y = x^2 + 1 is a parabola which is symmetric about the Y axis

bnt

That looks like it - nice!

Dr.Millah

bnt wrote: »

The limits are right (-3, 2), and so is the idea, but there's a problem with your integration, because the values are off. Hint: integral of x is (x^2)/2, integral of x^2 is (x^3)/3 .

I just edited my post i forgot to add the first part of the integral when i was doing the limits (rushing) i got 37.5 for the area under the line.

bnt

Dr.Millah wrote: »

I just edited my post i forgot to add the first part of the integral when i was doing the limits (rushing) i got 37.5 for the area under the line.

Ah - I looked again, thought I'd misread the answer, and edited what I wrote. Well, it's sorted.

Dr.Millah

Ha! Editing can be pretty confusing.

Thanks for the help everyone. Much appreciated

sponsoredwalk

This may be a little late but it reminds me of this video,

http://www.youtube.com/watch?v=upO6Mh862PI&feature=youtube_gdata

hopefully you didn't understand it fully so my amazing video will come to the rescue lol.

Grudaire

sponsoredwalk wrote: »

This may be a little late but it reminds me of this video,

http://www.youtube.com/watch?v=upO6Mh862PI&feature=youtube_gdata

hopefully you didn't understand it fully so my amazing video will come to the rescue lol.

8 minutes!? Jaysus, if he took the time to notice that the area of the sin curve 'cancels' we can write the integral as:



(which can be solved in 4 lines)I think the key is drawing it out.

Very good vid btw, it actually makes sense