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# Maths Homework from Unit 2 people..

### #1

Posted 07 December 2006 - 09:09 PM

Q1. Find the value of " integrate sign with 4 on top and 1 on bottom" (sorry) Square root of x then dx

HELP would be fantastic.

I'd like to understand.

x

### #2

Posted 07 December 2006 - 09:15 PM

root x is the same as x^1/2.

So... To integrate, add 1 to the power and divide by the power. giving (x^3/2) all over 3/2. Which is the same as (root x^3) all over 3/2. You then turn the denominator upside down to give 2(root x^3/2) all over 3.

P.s. (If someone could put this in an image it would really help)

This helping at all?

### #3

Posted 07 December 2006 - 09:17 PM

Then you will perform a standard integration (raise the power and divide by new power)

After that you will subsitute the x value with the upper limit (the top value) and and do the same with the lower limit (the bottom number)

And then you will get your answer by subtracting the second answer(we used the lower limit for this one) from the answer gained from the first substitution(we used the upper limit)

EDIT: Beaten to it! I'll try and get an image though asap.

EDIT 2: No need for picture lol

### #4

Posted 07 December 2006 - 09:18 PM

What you do is convert the square root around the expression to a familiar fractional index form. Remembering the laws of indices, a square root would be x to the power of 1/2. You can then integrate that by adding one to the power, and then dividing by the new power, like so:

What you then do, is substitute the upper and lower limits for x. You subtract the higher limit from the lower limit.

Hope you're following me alright?

Edit: Meh, you beat me to it, John.

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### #5

Posted 07 December 2006 - 09:19 PM

Right, so you want to evaluate:

What you do is convert the square root around the expression to a familiar fractional index form. Remembering the laws of indices, a square root would be x to the power of 1/2. You can then integrate that by adding one to the power, and then dividing by the new power, like so:

What you then do, is substitute the upper and lower limits for x. You subtract the higher limit from the lower limit.

Hope you're following me alright?

That in english would look nice Wedge

### #6

Posted 07 December 2006 - 09:21 PM

Right, so you want to evaluate:

What you do is convert the square root around the expression to a familiar fractional index form. Remembering the laws of indices, a square root would be x to the power of 1/2. You can then integrate that by adding one to the power, and then dividing by the new power, like so:

What you then do, is substitute the upper and lower limits for x. You subtract the higher limit from the lower limit.

Hope you're following me alright?

That in english would look nice Wedge

I already fixed it before you posted that.

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### #7

Posted 07 December 2006 - 09:21 PM

STILL WON THO!

### #8

Posted 07 December 2006 - 09:26 PM

I don't understand how we got the 2 in front of it all though? sorry

xx

### #9

Posted 07 December 2006 - 09:28 PM

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### #10

Posted 07 December 2006 - 09:30 PM

GAH wedge sucks

### #11

Posted 07 December 2006 - 09:30 PM

It's jsut no ma subject at all.

x

### #13

Posted 07 December 2006 - 09:32 PM

x

I REALLY REALLY REALLY APPRECIATE THIS

### #14

Posted 07 December 2006 - 09:34 PM

or simply put, raise the power by one and divide by that new power.

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### #15

Posted 07 December 2006 - 09:34 PM

Giving you 2(4^3) all over 3

### #16

Posted 07 December 2006 - 09:37 PM

Giving you 2(4^3) all over 3 smile.gif

You forgot to find the root of 4 there, will.

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### #17

Posted 07 December 2006 - 09:38 PM

Edit: bit dissapointed with myself for that...

### #18

Posted 07 December 2006 - 09:39 PM

and if it is? how do i do that?

x

### #19

Posted 07 December 2006 - 09:40 PM

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### #20

Posted 07 December 2006 - 09:41 PM

If you do square root of 4^3 its a bit easier.

BTW square root of 4^3 = 4^3/2

EDIT: RIGHT WEDGE...ITS ON! lol

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