Integration

OP

Original post by notnek

The method you are using is to find the "area enclosed by a curve and a line".

This region is enclosed by the curve, the line and the x-axis. It is not only enclosed by the curve and the line.

So the method would not work here.

Oh - I guess I thought I took the x axis into account because of the limits, 10/3 and 0

So for future reference if it involves the x axis dont use this method? Just want to pick the right one in an exam!

Reply 3

Original post by christinajane

Oh - I guess I thought I took the x axis into account because of the limits, 10/3 and 0

So for future reference if it involves the x axis dont use this method? Just want to pick the right one in an exam!

I'm confused, i thought this method included the x axis, let me work through the question

Reply 4

OP

Original post by LewisClothier

I'm confused, i thought this method included the x axis, let me work through the question

Yeah I am still a bit - I thought it did too - unless ive calculated something else wrong....

Reply 5

Original post by christinajane

Oh - I guess I thought I took the x axis into account because of the limits, 10/3 and 0

So for future reference if it involves the x axis dont use this method? Just want to pick the right one in an exam!

did you just integrate under the curve? or did you integrate with the line - the curve?

when you have two lines or a curve and a line you need to take the one that is underneath away from the one on top then integrate, It's difficult to explain

Reply 6

OP

Original post by LewisClothier

did you just integrate under the curve? or did you integrate with the line - the curve?

when you have two lines or a curve and a line you need to take the one that is underneath away from the one on top then integrate, It's difficult to explain

I done the curve minus the line, because the curve was over the line and then integrated the resulting equation...

(edited 7 years ago)

Reply 7

Original post by christinajane

Yeah I am still a bit - I thought it did too - unless ive calculated something else wrong....

yeah i just did it how i usually would and didn't get the answer on the mark scheme...

Reply 8

OP

Original post by LewisClothier

yeah i just did it how i usually would and didn't get the answer on the mark scheme...

Yeah weird - Its driving me insane

Reply 9

Original post by christinajane

I done the curve minus the line, because the curve was over the line and then integrated the resulting equation...

pretty sure you have to do the line - the curve but it obviously just comes out the same but positive, i don't know here, sorry

Reply 10

Original post by christinajane

Yeah weird - Its driving me insane

did you get 1900/81 ?

Reply 11

OP

Yeah no worries - maybe notnek is right but it seems to make sense what we did....

Because I have done the same thing in the past with other questions and its worked...

Reply 12

OP

Original post by LewisClothier

did you get 1900/81 ?

Yeah it was something over 81 - not sure now coz i threw my paper in the bin in frustration.

Will have another go at it in a lil bit - will probably get something completely different now!

Reply 13

Original post by christinajane

Yeah it was something over 81 - not sure now coz i threw my paper in the bin in frustration.

Will have another go at it in a lil bit - will probably get something completely different now!

I can't even work out what they've done on the mark scheme, can you?

Reply 14

https://www.youtube.com/watch?v=CRTYcP80GdE

Original post by christinajane

Yeah it was something over 81 - not sure now coz i threw my paper in the bin in frustration.

Will have another go at it in a lil bit - will probably get something completely different now!

Reply 15

OP

https://www.youtube.com/watch?v=CRTYcP80GdE

I dont even know how they decipher mark schemes in general!

But find the area of the triangle and then add that to the integrated bit between the limits of 0 and 2.

I know I should jsut stick with that one - but its just annoying me that this method - which should work, doesnt give the right answer.

Reply 16

Zacken

22

Original post by LewisClothier

Original post by christinajane

I dont even know how they decipher mark schemes in general!

But find the area of the triangle and then add that to the integrated bit between the limits of 0 and 2.

I know I should jsut stick with that one - but its just annoying me that this method - which should work, doesnt give the right answer.

The area that you have found is the blue area plus the green area (they will have opposite signs and cancel each other out a little).

As you can see, the area that you have found (the area between the curve and the line) is nothing close to the area that you need to find.

Reply 17

Notnek

21

Original post by christinajane

Yeah weird - Its driving me insane

It won't work here as I said. Think about what

The method you are using is to find the "area enclosed by a curve and a line".

This region is enclosed by the curve, the line and the x-axis. It is not only enclosed by the curve and the line.

So the method would not work here.

Just adding to this, you can also use the method to find the area enclosed by two curves and two vertical lines

x=a

and

x=b

.

Reply 18

Notnek

21

Original post by christinajane

Why does the subtraction method work:

\displaystyle \int^1_0 \sqrt{x}-x^2 \ dx = \int^1_0 \sqrt{x} \ dx - \int^1_0 x^2 \ dx

So what you're actually doing is subtracting the area under the red line from the area under the blue line. You should be able to see that this gives the green area.

Now in your question, what you have done is subtract the area under the line between 0 and N from the area under the curve between 0 and N. Look at your diagram and think about why that won't give you the required area.

Reply 19

OP