You are Here: Home >< Maths

C2 intergration!!! Watch

1. y= x^2/3 - 2/x^1/3 + 1 point A (8,4) lies on this curve and the curve crosses the x-axis at (1,0) the point B is (4,0) The line passes through AB. Find the equation of the line which I got to be y=x-4 The finite region R is bounded by C, AB and the positive X axis . Find the area of R. I intergrated the line- curve which I got to be 1/2x^2 -4x-3/5x^5/3+3x^2/3-x I intergrated using the points 8 ad 1 which I think is wrong. Any ideas?? the answer is supposed to be 43/5.
2. (Original post by SophieL1996)
...
Refering to the attached diagram.

The curve is above the line, so you'd want "curve" - "line".

With that correction, you'd get the whole shaded area.

But you only want the crosshatched area, so you need to subtract the area in red only.

Alternatively, you could work out the area from x=1 to x=4 above the x-axis, and then add on the difference between the curve and the line between x=4 and x=8.

Motto: Do a sketch.
Attached Images

3. (Original post by SophieL1996)
y= x^2/3 - 2/x^1/3 + 1 point A (8,4) lies on this curve and the curve crosses the x-axis at (1,0) the point B is (4,0) The line passes through AB. Find the equation of the line which I got to be y=x-4 The finite region R is bounded by C, AB and the positive X axis . Find the area of R. I intergrated the line- curve which I got to be 1/2x^2 -4x-3/5x^5/3+3x^2/3-x I intergrated using the points 8 ad 1 which I think is wrong. Any ideas?? the answer is supposed to be 43/5.
I'm struggling to sort out what your function looks like!

Is it supposed to be

or something different?

Does the question say the line passes through AB or the line passes through A and B or something else? I'm not sure how the positive x-axis comes in if you just need the area between C and the line
4. (Original post by ghostwalker)
Refering to the attached diagram.

The curve is above the line, so you'd want "curve" - "line".

With that correction, you'd get the whole shaded area.

But you only want the crosshatched area, so you need to subtract the area in red only.

Alternatively, you could work out the area from x=1 to x=4 above the x-axis, and then add on the difference between the curve and the line between x=4 and x=8.

Motto: Do a sketch.
Excellent diagram. I have used it to confirm the answer is indeed 43/5 - let's hope the OP can get it too
5. (Original post by davros)
Excellent diagram. I have used it to confirm the answer is indeed 43/5 - let's hope the OP can get it too
sorry I am still confused... could you show me some of your working to get me started please? and in the diagram shouldn't the curve be a parabola ?
6. y= x^2/3 - 2/x^1/3 + 1 point A (8,4) lies on this curve and the curve crosses the x-axis at (1,0) the point B is (4,0) The line passes through AB. Find the equation of the line which I got to be y=x-4 The finite region R is bounded by C, AB and the positive X axis . Find the area of R. I intergrated the line- curve which I got to be 1/2x^2 -4x-3/5x^5/3+3x^2/3-x I intergrated using the points 8 ad 1 which I think is wrong. Any ideas?? the answer is supposed to be 43/5.
7. (Original post by ghostwalker)
Refering to the attached diagram.

The curve is above the line, so you'd want "curve" - "line".

With that correction, you'd get the whole shaded area.

But you only want the crosshatched area, so you need to subtract the area in red only.

Alternatively, you could work out the area from x=1 to x=4 above the x-axis, and then add on the difference between the curve and the line between x=4 and x=8.

Motto: Do a sketch.
Thankyou for the diagram it explains a lot How come the area under the curve from 0 to 1 is not included in the shaded area ?
8. (Original post by SophieL1996)
y= x^2/3 - 2/x^1/3 + 1 point A (8,4) lies on this curve and the curve crosses the x-axis at (1,0) the point B is (4,0) The line passes through AB. Find the equation of the line which I got to be y=x-4 The finite region R is bounded by C, AB and the positive X axis . Find the area of R. I intergrated the line- curve which I got to be 1/2x^2 -4x-3/5x^5/3+3x^2/3-x I intergrated using the points 8 ad 1 which I think is wrong. Any ideas?? the answer is supposed to be 43/5.
I'm assuming they haven't given you a sketch. So, to aid your understanding, you could sketch the curve C using it's key features, with the line on the same set of axes.

Note:

- asymptotes
- points where it crosses the axes.
- behaviour when x is small
- behaviour when x is large

They will not give you such a horrible question without a sketch in the exam.
9. (Original post by Indeterminate)
I'm assuming they haven't given you a sketch. So, to aid your understanding, you could sketch the curve C using it's key features, with the line on the same set of axes.

Note:

- asymptotes
- points where it crosses the axes.
- behaviour when x is small
- behaviour when x is large

They will not give you such a horrible question without a sketch in the exam.
Ok thanks I have a rough sketch but I have no idea how to find the area ..
10. (Original post by SophieL1996)
y= x^2/3 - 2/x^1/3 + 1 point A (8,4) lies on this curve and the curve crosses the x-axis at (1,0) the point B is (4,0) The line passes through AB. Find the equation of the line which I got to be y=x-4 The finite region R is bounded by C, AB and the positive X axis . Find the area of R. I intergrated the line- curve which I got to be 1/2x^2 -4x-3/5x^5/3+3x^2/3-x I intergrated using the points 8 ad 1 which I think is wrong. Any ideas?? the answer is supposed to be 43/5.
Integrate only C from 1 to 8
and subtract the area under the line (from 4 to 8 triangle) from this
11. (Original post by SophieL1996)
sorry I am still confused... could you show me some of your working to get me started please? and in the diagram shouldn't the curve be a parabola ?
It's not a parabola because you have fractional powers of x (2/3 and -1/3 in there), and as x gets close to 0 the term becomes infinite!

I tackled this the same as the last question you posted - in other words I integrated from x=1 to x=8 to get the total area under the curve for that range and then subtracted the area of the triangle with vertices at (4,0), (8,0) and (8,4), i.e. the triangle formed by the x-axis and the lines y=x-4 and x=8.

What do you get when you integrate with respect to x?

(Original post by SophieL1996)
Thankyou for the diagram it explains a lot How come the area under the curve from 0 to 1 is not included in the shaded area ?
The question says the area is bounded by the positive x-axis, so you start your calculation from the point x=1.
12. (Original post by SophieL1996)
y= x^2/3 - 2/x^1/3 + 1 point A (8,4) lies on this curve and the curve crosses the x-axis at (1,0) the point B is (4,0) The line passes through AB. Find the equation of the line which I got to be y=x-4 The finite region R is bounded by C, AB and the positive X axis . Find the area of R. I intergrated the line- curve which I got to be 1/2x^2 -4x-3/5x^5/3+3x^2/3-x I intergrated using the points 8 ad 1 which I think is wrong. Any ideas?? the answer is supposed to be 43/5.
Why have you double-posted this question? I think your integration is right but it looks like you've subtracted the curve equation from the line, rather than the other way round What do you get when you put in the limits of 8 and 1?
13. Thanks everyone I got the correct answer!
14. (Original post by SophieL1996)
Thanks everyone I got the correct answer!
well done

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 3, 2013
Today on TSR

'Entry requirements are a form of elitism'

What do you think?

Roommate kissed my boyfriend

Discussions on TSR

• Latest
• See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

How to use LaTex

Writing equations the easy way

Study habits of A* students

Top tips from students who have already aced their exams

Chat with other maths applicants

Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE