Intergration Questions (1 Viewer)

[Avenger6]

Hi could someone help me/explain to me how I can solve this:

Find the area bounded by the curve y=9-x^2 and the line y=5.

Any help is appreciated .

 

[STx]

First try drawing the two curves, then decide which one of them is greater, their intercept points then find the integral of the larger curve minus the smaller one and let the bounds be from the two intercept points.

 

[Avenger6]

Ok, I still dont really understand how I can solve this as y=5 is not a curve. Im unsure of what area I have to measure.

 

[midifile]

Ok, I still dont really understand how I can solve this as y=5 is not a curve. Im unsure of what area I have to measure.

Even though y=5 is not a curve it is still a line, so in one graph draw y=9-x^2 and a horizontal line through 5 on the y-axis (y=5).

Then you can work out where they intersect by solving them simultaneously:
y=5 and y=9-x^2, so
5=9-x^2
x^2=4
x=+-2
therefore they intersect at x=2 and x=-2

Now all you do is do the integral from 2 to -2 of (9 - x^2 - 5) (as y=9-x^2 is the top curve).

Becuase the graph is symmetrical you can do 2 times the integral from 0 to 2 of (4 - x^2)

hope that helps

 

[cwag]

I have attached my answer...see if you can decipher.

 

[Aerath]

Here's the graph:

Edit: Seems someone else beat me to it.

 

[Avenger6]

Thanks guys i am understanding how it works now. However I do have another one which I am stuck on:

Find the area enclosed between the curve y=x^3, the x-axis and the line y=-3x+4.

So far I have come to x^3+3x-4=0 but have no idea how to factorise from here?

 

[Aerath]

You don't factorise it, you integrate from 0 to 4/3 for x^3 + 3x - 4, which should give you (x^4)/4 + (3x^2)/2 - 4x, with borders 0 and 4/3.

 

[Avenger6]

Yes but how did you find out what the border are. I can only seem to get x=1 as a border after drawing a the line and curve and am unsure of the other?

 

[Aerath]

Well, sketch it:

 

[Avenger6]

Ah I see know, i was looking at the wrong approach to get the boundaries. One issue now, when I integrate (x^4)/4 + (3x^2)/2 - 4x, with borders 0 and 4/3 I get an incorrect answer??? I should get 0.42 units^2 as an answer.

 

[Aerath]

You don't integrate (x^4)/4 + (3x^2)/2 - 4x, you sub the borders in. I've already integrated it for you.

 

[Avenger6]

Ah ok i finnally got it. I had to do it in two parts, I figured out where the line y=-3x+4 and curve y=x^3 intercepted which was at 1. Then I found the area of each seperatly i.e. y=x^3 with borders 0 and 1 and y=-3x+4 with borders 1 and 4/3. Then when both areas were added together I got 0.42units^2. Thanks for all the help, really bettered my understanding. Just out of curiosity though, should I use a diagram to work out the borders or use factorisation to work them out when doing these sort of questions?

 

[Aerath]

I always prefer to have a diagram - however, there is no set rules to do these things. Whatever floats your boat, I guess.

 

[Mark576]

Ah ok i finnally got it. I had to do it in two parts, I figured out where the line y=-3x+4 and curve y=x^3 intercepted which was at 1. Then I found the area of each seperatly i.e. y=x^3 with borders 0 and 1 and y=-3x+4 with borders 1 and 4/3. Then when both areas were added together I got 0.42units^2. Thanks for all the help, really bettered my understanding. Just out of curiosity though, should I use a diagram to work out the borders or use factorisation to work them out when doing these sort of questions?

Best to use a diagram if you're unsure.

Oh, and Aerath, you were doing it incorrectly, not sure what you were thinking, maybe you should wait until you do integration properly in school (unless you've already done it? But it doesn't seem like you have )?. Avenger6 is correct in his method.

 

[Aerath]

Wait, I thought you could integrate both things at the same time if you minused the bottom curve from the top curve?

If I was wrong, sorry for any confusion.

 

[tacogym27101990]

hey how ya's doin?
can someone help me with an integration question from the 2006 hsc?

It's finding the volume between the curves y=x^2 +1 and y+5 in the first quadrant only

i keep getting 8pi but my teacher says its 112pi/3.

 

[tacogym27101990]

around the y-axis sorry

 

[tacogym27101990]

yeah thats the one
and where i wrote y+5 i meant y=5.
thanks

and can someone please help me tonight if it's possible, its just i need it for tomoro. thanks

 

[tacogym27101990]

turns out my teacher was wrong and it was 8pi =]