for question 8, i can't understand how to do this! anytime i think i may have it, its always wrong! lol

same for 10, im just getting rly frustrated cosi cant get the right answer! thx

also sorry that i keep posting questions but im panicking before my higher

**0**

# 2003 Paper I

Started by kr88, Apr 18 2005 10:34 AM

4 replies to this topic

### #1

Posted 18 April 2005 - 10:34 AM

### #2

Posted 18 April 2005 - 10:48 AM

Question 8 is just standrad integration. The maths notes on Integration have similiar questions to this.

Remember the general formula for iintegration is:

x

How far do you get with this question?

Remember the general formula for iintegration is:

x

^{n}= (x^{n + 1})/ (*n*+ 1) + C.How far do you get with this question?

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### #3

Posted 18 April 2005 - 11:13 AM

thw working i have done so far is:-

= 1/(3x+1)^(1/2) dx =(3x+1)^-(1/2) dx = 3x^-(1/2) +1 dx

= [3x^(1/2) / (1/2) +X] (with limits 1 and 0)

= (3+1 / (1/2) +1) - (0)

= 4/(1/2) +1 =8/1 +1 =9

i know how to do integration and ewverything, and can do pastpaper integration questions that are more difficult than this, but something is wrong with my working as this is the wrong answer! lol sorry bout this

= 1/(3x+1)^(1/2) dx =(3x+1)^-(1/2) dx = 3x^-(1/2) +1 dx

= [3x^(1/2) / (1/2) +X] (with limits 1 and 0)

= (3+1 / (1/2) +1) - (0)

= 4/(1/2) +1 =8/1 +1 =9

i know how to do integration and ewverything, and can do pastpaper integration questions that are more difficult than this, but something is wrong with my working as this is the wrong answer! lol sorry bout this

### #4

Posted 18 April 2005 - 11:43 AM

No problem.

=

Now when you integrate this you keep '1' in the brackets so you get:

(3x + 1)

= [2 (3x + 1)

= [ 4/3 ] - [ 2/3]

=

Again if you don't understand any steps, post again.

The formula used this time is:

(ax + b)

In this example we just done a was

_{0}^{1}dx / (3x + 1)^{1/2}dx=

_{0}^{1}(3x + 1)^{-1/2}dxNow when you integrate this you keep '1' in the brackets so you get:

(3x + 1)

^{1/2}/ 3. (1/2)= [2 (3x + 1)

^{1/2}/3]_{0}^{1}= [ 4/3 ] - [ 2/3]

=

**2/3**.Again if you don't understand any steps, post again.

The formula used this time is:

(ax + b)

^{n}= (ax+b)^{n + 1}/ a .(*n*+ 1) + C.In this example we just done a was

**3**.
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### #5

Posted 18 April 2005 - 09:25 PM

ok thanks ur immense¬!

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