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secondary 4 | A Maths
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No option for polytechnic math, so I put this under Sec 4 A Math. Please help guide me on how to solve this integral. Thanks.
Date Posted: 3 years ago
Views: 495
This integral is of the form
f'(x) times f(x)
where f(x) ie e^(x^2) (if you realise, differentiating e will get you the same thing again, along with the derivative of the power).
Differentiating x^2 gives us 2x, which is basically what we need here (the 2 or 1/2 multiplier is easy to take care of later on).
f'(x) times f(x)
where f(x) ie e^(x^2) (if you realise, differentiating e will get you the same thing again, along with the derivative of the power).
Differentiating x^2 gives us 2x, which is basically what we need here (the 2 or 1/2 multiplier is easy to take care of later on).
Thanks for the explanation
See 1 Answer
Oh ok, thanks. But is there detailed steps to this qn? Because it is 4 marks, I don't think giving a short answer like this will help.
Is this the whole question. It will be nice if can see the whole question.
By the way, i did skip one or 2 steps.
Have you learnt integration by parts? If yes I can show you a solution employing its use
It just says find the following integral, that's all. No other useful info.
To J: Yes I've learnt that. Please show me, thanks for the help.
I doubt it can be done by using by-parts
I think the question may want you to initiate the derivative first then do the inverse differentiation concept.
To Yong: I dunno anything, this qn is so hard oh my gosh. I can't even figure out the first step.
Have you learnt integration by substitution? That would definitely be possible
let u = x²
Then du/dx = 2x
→"du = 2x dx "
∫ xeˣ² dx
= ∫ (½ eˣ²) 2x dx
= ∫ ½ eᵘ du
= ½ eᵘ
= ½eˣ² + c
(Change the u back to x²)
Then du/dx = 2x
→"du = 2x dx "
∫ xeˣ² dx
= ∫ (½ eˣ²) 2x dx
= ∫ ½ eᵘ du
= ½ eᵘ
= ½eˣ² + c
(Change the u back to x²)
Alternatively, this is directly integrable so just write this :
∫ xeˣ² dx
= ½ ∫ 2xeˣ² dx
= ½ ∫ (d/dx (eˣ²) ) dx
= ½ eˣ² + c
∫ xeˣ² dx
= ½ ∫ 2xeˣ² dx
= ½ ∫ (d/dx (eˣ²) ) dx
= ½ eˣ² + c
To Yong : it can be done, but it is quite advanced.
https://math.stackexchange.com/questions/2217850/integration-by-parts-int-xex2-dx
https://math.stackexchange.com/questions/2217850/integration-by-parts-int-xex2-dx
Ughhhh, alright thank you yall, both Yong and J
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