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intergration question

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Original post by djmans
if its f'(x) what should i do?


If it's asking you to find f'(x) that means differentiate it.
If it gives you f'(x) and it asks you to find the curve, then integrate it.




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B_9710
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Original post by djmans
so f its f'(x) i must integrate


f'(x) is the derivative with repsect to x of f(x) so to find f(x) you must 'undo' the differentiation, so you must integrate.
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Original post by Zacken


could you help me with this (b) its an easy question but i am struggling to get the answers.
Capture.PNG24.0KB
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Zacken
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Original post by djmans
could you help me with this (b) its an easy question but i am struggling to get the answers.


Re-arrange one of the equations in the form y = mx + c and then sub this into the other equation.
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Original post by Zacken
Re-arrange one of the equations in the form y = mx + c and then sub this into the other equation.


that what i been trying for than last hour but cant get the answer, could you please help:smile:
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Zacken
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Original post by djmans
that what i been trying for than last hour but cant get the answer, could you please help:smile:


Put up a picture of your working.
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Original post by Zacken
Put up a picture of your working.


i cant upload it, its too big
Reply 27
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Zacken
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Original post by djmans
i cant upload it, its too big


Line L1: m=12−311−−1=912=34m = \frac{12-3}{11--1} = \frac{9}{12} = \frac{3}{4}, so y−12=34(x−11)⇒y=12+34(x−11)y -12 = \frac{3}{4}(x-11) \Rightarrow y = 12 + \frac{3}{4}(x-11).

Plug this into the other equation:

3(34(x−11)+12)+4x=303\left(\frac{3}{4}(x-11) + 12\right) + 4x = 30 and solve this.
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djmans
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Original post by Zacken
Line L1: m=12−311−−1=912=34m = \frac{12-3}{11--1} = \frac{9}{12} = \frac{3}{4}, so y−12=34(x−11)⇒y=12+34(x−11)y -12 = \frac{3}{4}(x-11) \Rightarrow y = 12 + \frac{3}{4}(x-11).

Plug this into the other equation:

3(34(x−11)+12)+4x=303\left(\frac{3}{4}(x-11) + 12\right) + 4x = 30 and solve this.


thanks i got the answer but before i was trying the same method by rearranging the first equation and substituting into eq 2 but got something completely different
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Zacken
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Original post by djmans
thanks i got the answer but before i was trying the same method by rearranging the first equation and substituting into eq 2 but got something completely different


See if you can spot your mistake yourself, otherwise you'll need to put up your working in some way for me to check. But, make sure you understand what I did.
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Original post by Zacken
See if you can spot your mistake yourself, otherwise you'll need to put up your working in some way for me to check. But, make sure you understand what I did.

why did you sub directly using this eq from (a)

than using this eq, its the same thing but you get a different answer.
4y – 3x –15 = 0
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Original post by djmans
why did you sub directly using this eq from (a)

than using this eq, its the same thing but you get a different answer.
4y – 3x –15 = 0


No, you don't get a different answer; you're just doing it wrong somehow, but I can't check unless you show me your working.
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Original post by Zacken
No, you don't get a different answer; you're just doing it wrong somehow, but I can't check unless you show me your working.


you were right, i tried the question again today i got it.
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Zacken
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Original post by djmans
you were right, i tried the question again today i got it.


Awesome.

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