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Apr 29th, 2006, 03:13 PM
#1
Thread Starter
New Member
help needed
Hi, i am new to this forum and could do with some help on a question on intergration.
3
/ (3x(power of 5)-2x-3)dx
2
sorry about the format, any help appreciated.
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Apr 29th, 2006, 03:26 PM
#2
Member
Re: help needed
∫ 3x^5 -2x -3 dx
= (x^6)/2 –x2 -3x.
So with limits of 2 and 3, sub x=2 and take that from what you get subbing x=3.
= (346.5) - (22)
=324.5
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Apr 29th, 2006, 03:30 PM
#3
Re: help needed
ok i am denoting the integration sign with limits like 3/2
Code:
3/2 (3x^5 - 2x - 3) dx
= 3/2 (3x^5)dx - 3/2 (2x)dx - 3/2 (3)dx
= 3/2 (3(x^6)/6) - 3/2 (2(x^2)/2) - 3/2 (3x)
= ((3^6)/2 - (2^6)/2) - (3^2 - 2^2) - (3*3 - 3*2)
= (364.5 - 32) - ( 9 - 4 ) - (9 - 6)
= 332.5 - 5 - 3
= 324.5
sorry, but i think i am wrong. please someone correct me
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Apr 29th, 2006, 03:31 PM
#4
Re: help needed
hey thomas, how you made the integral sign??
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Apr 29th, 2006, 03:33 PM
#5
Member
Re: help needed
I copied it from word.
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Apr 29th, 2006, 03:37 PM
#6
Re: help needed
 Originally Posted by Thomas154321
I copied it from word. 
:slaps forehead:
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Apr 30th, 2006, 02:50 AM
#7
Thread Starter
New Member
Re: help needed
thanks guys kinda understand it, but i seem to be getting a different answer, here is a simpler one, maybe i will be able to follow your workings better. Sorry if i seem a bit slow, new starter you see,
3
/ x dx
2
thanks.
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Apr 30th, 2006, 03:59 AM
#8
Member
Re: help needed
To integrate the sum of terms, take each part separately. So first, take the 3x^5.
1) Add 1 to the power = 3x^6.
2) Divide by the new power = (3x^6)/6 = (x^6)/2.
3) ∫(-2x) dx = (-2x^2)/2 = -x^2.
4) ∫(-3) dx = -3x.
So, add them up to get (x^6)/2 -x^2 -3x
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Apr 30th, 2006, 11:43 AM
#9
Thread Starter
New Member
Re: help needed
Hi, think i have grasped it,
would i be right in saying that,
3
/ x dx would be, (3^2/2) - (2^2/2) = 9/2 - 4/2 = 4.5-2 =2.5.
2
thanks astro.
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Apr 30th, 2006, 11:45 AM
#10
Member
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Apr 30th, 2006, 12:39 PM
#11
Thread Starter
New Member
Re: help needed
thanks for all your help thomas and harsh gupta, could please re assure me on this last question, thanks,
2
/ (4x^2 - 5x + 2) dx
-1
(4x^3/3 - 5x^2/2 +2x)
(x^3/0.75 - x^2/2.5 + 2x)
(2^3/0.75 - -1^3/0.75) - (2^2/2.5 - -1^2/2.5) + (2*2 - 2*-1)
(10.7 - -1.3 - 1.6 - 0.4 + 6) = 16
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Apr 30th, 2006, 12:57 PM
#12
Member
Re: help needed
2
∫ (4x^2 - 5x + 2) dx =
-1
[(4/3)x^3 - (5/2)x^2 + 2x] from 2 to -1 =
((32/3)-10+4) - (-(4/3)-(5/2)-2) =
(14/3)-(-105/18) = 10.5.
You divided the middle term by 2.5 instead of multiplied. You did the basic integration correctly though, which is the hard part.
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Apr 30th, 2006, 01:21 PM
#13
Thread Starter
New Member
Re: help needed
sorry thomas could you simplify your workings for me, sorry if i seem a bit slow. thanks astro
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Apr 30th, 2006, 01:26 PM
#14
Member
Re: help needed
Ok, to integrate 4x^2 - 5x + 2, the 4x^2 goes to (4x^3)/3, the -5x goes to (-5x^2)/2, and the +2 goes to +2x.
Add these 3 to get (4x^3)/3 (-5x^2)/2 +2x.
Replace x by 2 and get a number, and then replace x by (-1) to get a second number. Take the second away from the first to get the answer.
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Apr 30th, 2006, 01:31 PM
#15
Thread Starter
New Member
Re: help needed
thanks for all your help thomas, its much appreciated. Its finally clicked now
regards astro
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Apr 30th, 2006, 01:53 PM
#16
Member
Re: help needed
Pleasure.
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