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
0
Maths Homework from Unit 2 people..
Started by lynnie, Dec 07 2006 09:09 PM
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#1
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Posted 07 December 2006 - 09:09 PM
Right I'm not the best at typing things in maths form.....So I'll do my best.
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
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
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Posted 07 December 2006 - 09:15 PM
Right...This is a nice and simple question.
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?
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
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Posted 07 December 2006 - 09:17 PM
First of all you must convert the equation into indice form
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
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
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Posted 07 December 2006 - 09:18 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:
![\begin{align*}\left[ {\frac{{2x^{\frac{3}{2}} }}{3}} \right]_1^4\end{align*}](/latexrender/pictures/35a39198cd38df56bc2c929345497869.gif "\begin{align*}\left[ {\frac{{2x^{\frac{3}{2}} }}{3}} \right]_1^4\end{align*}")
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.
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
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Posted 07 December 2006 - 09:19 PM
QUOTE(The Wedge Effect @ Dec 7 2006, 09:18 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
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Posted 07 December 2006 - 09:21 PM
QUOTE(will_789 @ Dec 7 2006, 09:19 PM)
QUOTE(The Wedge Effect @ Dec 7 2006, 09:18 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
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Posted 07 December 2006 - 09:21 PM
B*l*o*k* lol
STILL WON THO!
STILL WON THO!
#8
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Posted 07 December 2006 - 09:26 PM
Yeah! That kinna helps guys.
I don't understand how we got the 2 in front of it all though? sorry
xx
I don't understand how we got the 2 in front of it all though? sorry
xx
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Posted 07 December 2006 - 09:28 PM
The 2 in front of it all is because you're dividing by 3/2. When you divide something by that, you're effectively multiplying by the reciprocal, i.e, 2/3. Understand?
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#10
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Posted 07 December 2006 - 09:30 PM
Flip the fraction on the denominator upside down, and bring the number on top of the fraction to the top. Makes it much easy to solve 
GAH wedge sucks
GAH wedge sucks
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Posted 07 December 2006 - 09:30 PM
Yeah think I got it. I feel so silly btw.
It's jsut no ma subject at all.
x
It's jsut no ma subject at all.
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Posted 07 December 2006 - 09:32 PM
QUOTE(lynnie @ Dec 7 2006, 09:30 PM)
Yeah think I got it. I feel so silly btw.
It's jsut no ma subject at all.
x
Don't feel silly! We're here to help people out! And if you feel like you get stuck...Just ask! We all get stuck aswell
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Posted 07 December 2006 - 09:32 PM
I'm getting a bit stuck when substituting the 4.....:-s
x
I REALLY REALLY REALLY APPRECIATE THIS
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I REALLY REALLY REALLY APPRECIATE THIS
#14
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Posted 07 December 2006 - 09:34 PM
Good. Glad to hear that. All you need to remember for integration from unit two is the rule:
or simply put, raise the power by one and divide by that new power.
or simply put, raise the power by one and divide by that new power.
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#15
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Posted 07 December 2006 - 09:34 PM
Think it through. Just put the 4 in place of the x.
Giving you 2(4^3) all over 3
Giving you 2(4^3) all over 3
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Posted 07 December 2006 - 09:37 PM
Alright, just replace x by 4 in the first bit. You're simply finding the square-root cubed of 4, which is 8. The denominator in the power is a root, while the numerator, ie, the top part, is the power to which you put your number up to. Is that the bit you had trouble with? Just remember the laws of indices from SG/Int 2 maths, and you're sorted.
You forgot to find the root of 4 there, will.
QUOTE(will_789 @ Dec 7 2006, 09:34 PM)
Think it through. Just put the 4 in place of the x.
Giving you 2(4^3) all over 3 smile.gif
Giving you 2(4^3) all over 3 smile.gif
You forgot to find the root of 4 there, will.
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Posted 07 December 2006 - 09:38 PM
GAH i forgot the square root 
Edit: bit dissapointed with myself for that...
Edit: bit dissapointed with myself for that...
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Posted 07 December 2006 - 09:39 PM
I thought it would be 4 to the power of 3 over 2?? is it not?
and if it is? how do i do that?
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and if it is? how do i do that?
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#19
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Posted 07 December 2006 - 09:40 PM
Yes. 4 to the power of 3 over 2 is effectively the square root of 4, cubed. See above post for explanation.
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#20
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Posted 07 December 2006 - 09:41 PM
It is...BUT
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
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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