Integration.
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HELP ME!!!
I just don't get integration and I have an exam tmr!!
Teach me how to solve the problems!!
Oh, explain the formulas. -
Which one?
The obvious definition(and easiest to understand) is:
| f(x) dx = F(x) where d/dx [F(x)] = f(x) -
How about you post some of your questions here so I can help?
I need to know your "level". -
dude, exam is either mid year of end of year. Intergration is taught at lower secondary. I think you want to say test, instead of exams rite?Originally posted by Quinsy:HELP ME!!!
I just don't get integration and I have an exam tmr!!
Teach me how to solve the problems!!
Oh, explain the formulas. -
No, for Engineering Maths. I dropped out of A maths in sec school thats why.Originally posted by Darkness_hacker99:dude, exam is either mid year of end of year. Intergration is taught at lower secondary. I think you want to say test, instead of exams rite?
Integration level is like, sec sch level not even there...
INtegrate e^4x / 1+e^4x dx -
Okay, a question.
| e^4x/(1 + e^4x) dx
Let u = e^4x
--> du/dx = 4e^4x = 4u
--> dx = du/4u
Hence,
| [u/(1 + u)] du/4u
= 1/4 | 1/(1+u) du
= 1/4 ln (1 + u) + C where C is an arbitary constant
= 1/4 ln (1 + e^4x) + C (Note: You don't need to worry about absolute sign in ln (1 + e^4x) since you know that e^4x is always positive)
Tip: Whenever you see two similar "terms", try using the substitution method.
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all those calculus involving exponentials...i could NEVER do
well, I would like to forget all those...so I can't help. I could, maybe like a few months ago... -
how come got ln?! 
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Exponentials are pretty okay since their differentiation and integrals are quite nice.Originally posted by finoq:all those calculus involving exponentials...i could NEVER do
well, I would like to forget all those...so I can't help. I could, maybe like a few months ago...
It's just that if the power involves a lot of terms, then one has to be careful, something like exp(At + B) where A, B are real constants(can be complex) etc.
Of course, if they want to put something so "bird" as:
| e^x * sin x * ln x dx
That I don't know how to manage. -
You can use "log" ala base 10 or "ln" ala base e.Originally posted by Quinsy: how come got ln?!
It's all the same lah. Use whichever you are using now. -
Kayz...
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It's all about practicing makes perfect. Do more and sooner or later, you'll have the "feel" for it. Of course, in times of tests/exams, you need to pray that you can remember whatever you did.Originally posted by Quinsy:Kayz...
As the kid in "I not stupid" said, "Must "mo" all the "di", then you can learn to defeat it!" -
Integrate (3-x^2) ^3 2x dx
T.T -
Do you mean 2x*(3 - x^2)^3? (* is multiply)Originally posted by Quinsy:Integrate (3-x^2) ^3 2x dx
T.T
If so, then
| 2x*(3 - x^2)^3 dx
Again, you can see that d/dx (x^2) = 2x, so your first thought shall be try to use substitution method:
Let u = x^2 --> du/dx = 2x --> dx = du/2x
Therefore
| 2x*(3 - x^2)^3 dx
= | 2x*(3 - u)^3*(du/2x)
= | (3 - u)^3 du (The 2x cancelled each other away)
= - (1/4) (3 - u)^4 + C where C is an arbitrary constant
Don't forget to sub back in u = x^2
= - (1/4) (3 - x^2)^4 + C (ANS) -
T.T
Me integration CMI liao lah... -
Cannot say CMI. At least must try.Originally posted by Quinsy:T.T
Me integration CMI liao lah...
The most thing to note, even for myself, is to remain calm even though the question may look scary.
Remember, you are tested on what you have been taught and learned. The solution has to come from somewhere that you know.
Whenever you see a "tough" question, take a deep breath, mai kan cheong and look and study the question again.
Hope this helps. -
apparantly all the probs got to do with chain rule??
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O.oOriginally posted by hisoka:apparantly all the probs got to do with chain rule??
really? -
both questions hor, its use the substitution to solve one thats chain rule i think. not confirm though. but both questions are about the substitution methodOriginally posted by Quinsy:O.o
really?