x Turn on thread page Beta
 You are Here: Home >< Maths

# differentiation confused? watch

1. (Original post by kiiten)
Then i subbed in 3 and 1 separately then took them away to get 25?

So 14 x 3 - 1/3 × 3^2 + 9 × 3^-1 for the first term (3)
for 3 it should be
2. (Original post by DylanJ42)
for 3 it should be
Thanks, careless mistake

Posted from TSR Mobile
3. (Original post by kiiten)
Thanks, careless mistake

Posted from TSR Mobile
dw we all make them
4. (Original post by DylanJ42)
dw we all make them
Wait, that was a typo. I still get 36 then 11 when i sub in 1? (Answer: 25) :s

Posted from TSR Mobile
5. (Original post by kiiten)
Wait, that was a typo. I still get 36 then 11 when i sub in 1? (Answer: 25) :s

Posted from TSR Mobile
What seems to be the problem? Sorry I wasn't part of the conversation so I don't know what's going on!

Also: it would be good if you could post your working out, that way we may be able to pinpoint where you're making mistakes!
Furthermore, I sometimes find it helpful that when a question is getting out of hand and messy, start it again on a new piece of paper

This is where I think we've gotten up to so far:

So you need to find

Your integration is correct, we now have

First you substitute in the limits (3 and 1), then you subtract what you get for the lower limit from what you get with the upper limit, so you have

Which is the area enclosed between and :

When

This is correct, however:

When

As
6. (Original post by edothero)
What seems to be the problem? Sorry I wasn't part of the conversation so I don't know what's going on!

Also: it would be good if you could post your working out, that way we may be able to pinpoint where you're making mistakes!
Furthermore, I sometimes find it helpful that when a question is getting out of hand and messy, start it again on a new piece of paper

This is where I think we've gotten up to so far:

So you need to find

Your integration is correct, we now have

First you substitute in the limits (3 and 1), then you subtract what you get for the lower limit from what you get with the upper limit, so you have

Which is the area enclosed between and :

When

This is correct, however:

When

As
Yeah, i must have typed it wrong into my calculator. So i got 40/3 and the area has to = 5 1/3 so you have to minus (4x2). But where does the 2 come from?
7. (Original post by kiiten)
Yeah, i must have typed it wrong into my calculator. So i got 40/3 and the area has to = 5 1/3 so you have to minus (4x2). But where does the 2 come from?
Sorry I'm really busy right now. Look at the graph. Area under the curve from x= 1 to 3 is a rectangle + curve.
8. (Original post by kiiten)
Yeah, i must have typed it wrong into my calculator. So i got 40/3 and the area has to = 5 1/3 so you have to minus (4x2). But where does the 2 come from?
As edothehero said, the area you've found using integration is the area under the curve. But you want the shaded area so you want to subtract the area of that rectangle.
9. (Original post by Zacken)
As edothehero said, the area you've found using integration is the area under the curve. But you want the shaded area so you want to subtract the area of that rectangle.
As i said before you have to minus 8 (4x2) from 40/3 to get 5 1/3. But, where does the 2 come from?

Posted from TSR Mobile
10. (Original post by kiiten)
As i said before you have to minus 8 (4x2) from 40/3 to get 5 1/3. But, where does the 2 come from?

Posted from TSR Mobile
What's the area of the rectangle? Width * height, no...?
11. (Original post by Zacken)
What's the area of the rectangle? Width * height, no...?
Height is 4 and i dont know about the width - maybe 3? The question says that the shaded area has to equal 5 1/3 so i did 40/3 - 5 1/3 to get 8. Thats why i thought the width might be 2?
12. (Original post by kiiten)
Height is 4 and i dont know about the width - maybe 3? The question says that the shaded area has to equal 5 1/3 so i did 40/3 - 5 1/3 to get 8. Thats why i thought the width might be 2?
You're integrating from 1 to 3, the rectangle has width 3-1 = 2.
13. (Original post by Zacken)
You're integrating from 1 to 3, the rectangle has width 3-1 = 2.
Oh, i must have got confused from the diagram because the curve goes inwards at the top so i thought it would be less. But, i see what you mean. Thanks
14. (Original post by kiiten)
Oh, i must have got confused from the diagram because the curve goes inwards at the top so i thought it would be less. But, i see what you mean. Thanks
The curve is, the rectangle isn't.

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: March 2, 2016
Today on TSR

### Did he block me?

What should I do?

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

## Groups associated with this forum:

View associated groups

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