Integration (mathematical) - a bit of help needed

[Knusper]

Multiply it out.
x*(25-x)(25-x)(25-x)
= x*(25-x)(625-50x+x^2)
= x*(15625-1250x+25x^2-625x+50x^2-x^3)
=(15625-1875x+75x^2-x^3)*x
=15625x-1875x^2+75x^3-x^4

Integral should be 7812.5x^2-625x^3+18.74x^4-0.2x^5+c

I don't quite agree with the method. I just commissioned my maths-genius friend to figure it out and he said to binomially expand instead of multiplying out. I'll have the answer in a jiffy.

It's the most intuitive method, that's all. Definitely not the fastest, but I think it's easier to follow the thread of logic this way.

Yeah I just figured it out and got to where you are so you were right after all, sorry.

It's a pity because I have an exam on this on Monday and I clearly don't know anything about it.

Hey, no need to apologise! I'm glad you were able to figure it out! Good luck with the exam. Is it just on calculus or is it on a range of topics?

Just calculus - Binomial distribution, differentiation, integration, trig identities - all the good stuff

 

[Melon Hunter]

Multiply it out.
x*(25-x)(25-x)(25-x)
= x*(25-x)(625-50x+x^2)
= x*(15625-1250x+25x^2-625x+50x^2-x^3)
=(15625-1875x+75x^2-x^3)*x
=15625x-1875x^2+75x^3-x^4

Integral should be 7812.5x^2-625x^3+18.74x^4-0.2x^5+c

I don't quite agree with the method. I just commissioned my maths-genius friend to figure it out and he said to binomially expand instead of multiplying out. I'll have the answer in a jiffy.

It's the most intuitive method, that's all. Definitely not the fastest, but I think it's easier to follow the thread of logic this way.

Yeah I just figured it out and got to where you are so you were right after all, sorry.

It's a pity because I have an exam on this on Monday and I clearly don't know anything about it.

Hey, no need to apologise! I'm glad you were able to figure it out! Good luck with the exam. Is it just on calculus or is it on a range of topics?

Just calculus - Binomial distribution, differentiation, integration, trig identities - all the good stuff

Awesome. Have fun with that! On that note, I should probably revise maths as well seeing as I have a 3 hour exam next Tuesday...

 

[Lukeje]

you use u substitution

Spoiler: this is my explanation

u= 25-x
x=25-u

so then the function becomes (25-u)*u^3

from there you distribute the u^3 ---> 25u^3-u^4 and integrate

you will get (25/4)*u^4 - (1/5)*u^5

you then replace u with 25-x to get the real answer

(25/4)*(25-x)^4 - (1/5)*(25-x)^5

I'm a computer science major so I have to know how to do these things

edit:

or you could do the simpler way that everyone else is talking about. Why do I always over-complicate things!?!

*face-desk*

You forgot to take into account that du=-dx (so all your answers should be the negative of what they are now).

 

[Durananrananrananran]

Allright. So I have this mathematical problem right here that I'm not able to solve. Don't look at me that way, I'm an electrician, we don't usually integrate =P

...snip

more maths threads like this. Much better than the one that had everyone arguing about BODMAS ad nauseum.

 

[Pappytech]

This. Is. Awesome. I know where I'm looking next time I need help on a math assignment.

Also, are people on here good at physics?

 

[Daye.04]

more maths threads like this. Much better than the one that had everyone arguing about BODMAS ad nauseum.

Hey, I remember that one! That was fun to read, wasn't it? =P
Quite like the 0.999'=1 discussion =D

But that's noted. The next time I have a mathematical problem, I'll shove it up here =D

 

[Lukeje]

This. Is. Awesome. I know where I'm looking next time I need help on a math assignment.

Also, are people on here good at physics?

There are some people on here who are good at Physics. Unfortunately they tend to get drowned out by the majority who are not (as most often happens on the internet).

 

[Daye.04]

There are some people on here who are good at Physics. Unfortunately they tend to get drowned out by the majority who are not (as most often happens on the internet).

Well. it depends on how you define physics ... I mean I'm awesome at calculating the forces one a box sliding down a hill ^^

 

[Durananrananrananran]

This. Is. Awesome. I know where I'm looking next time I need help on a math assignment.

Also, are people on here good at physics?

I did it for a year at Cambridge, but that was years ago. I should still be ok at school level stuff, maybe.

Although, this being the internet, any correct response I give will probably get drowned out by people who think they know everything because they think they understand Newton's laws of motion

 

[Trivun]

you use u substitution

Spoiler: this is my explanation

u= 25-x
x=25-u

so then the function becomes (25-u)*u^3

from there you distribute the u^3 ---> 25u^3-u^4 and integrate

you will get (25/4)*u^4 - (1/5)*u^5

you then replace u with 25-x to get the real answer

(25/4)*(25-x)^4 - (1/5)*(25-x)^5

I'm a computer science major so I have to know how to do these things

edit:

or you could do the simpler way that everyone else is talking about. Why do I always over-complicate things!?!

*face-desk*

Well, I do Maths at university (not great at it, but then again I am a lazy sod ) - that is to say, I just finished the three year course I was doing - and I'd say substitution is the easier way. I would answer the question for the OP except there'd be no point as the correct answer and explanation has been posted more than a dozen times by this point, including in your post, dude I'm quoting . But yeah, substitution is much easier because then you don't need to waste time multiplying out brackets for no good reason. The only reason I would ever use that method instead, multiplying out and doing the standard integral, is if I was told I had to get the answer in terms of x rather than terms of (25-x) or whatever...

 

[JoshGod]

I used integration by parts and got
(X/4)(25-X)^4 + (1/20)(25-X)^5
Edit;
i suspect something is wrong as differntiation of that seems wierd

 

[Pappytech]

This. Is. Awesome. I know where I'm looking next time I need help on a math assignment.

Also, are people on here good at physics?

I did it for a year at Cambridge, but that was years ago. I should still be ok at school level stuff, maybe.

Although, this being the internet, any correct response I give will probably get drowned out by people who think they know everything because they think they understand Newton's laws of motion

Sigh. No kidding.

 

[Kaland]

Integration by parts is probably easiest for this.

@JoshGod: Almost. You just forgot a minus. I got int[x(25-x)^3]=-[(x/4)(25-x)^4+(1/20)(25-x)^5]+C
Expand that, and you get -(1/5)(x^5)+(3/4)(25*x^4)-(25^2*x^3)+(25^3/2)x^2+C which is what Wolfram said.

When derived that yields (1/4)(25-x)^3+x(25-x)^3-(1/4)(25-x)^4=x(25-x)^3

The tiresome thing about the exercise is expanding the brackets, which is avoided by doing integration by parts.

 

[jwadd2100]

You can use a combination of the product rule and the chain rule.
f(x)g(x) = f'(x)g(x) + f(x)g'(x)
d(x(25-x)^3)/dx = (1)(25-x)^3 + 3x(x-25)^2*(1).
Hope this helps.

 

[Kaland]

You can use a combination of the product rule and the chain rule.
f(x)g(x) = f'(x)g(x) + f(x)g'(x)
d(x(25-x)^3)/dx = (1)(25-x)^3 + 3x(x-25)^2*(1).
Hope this helps.

He didn't ask how to differentiate f(x), he wanted to integrate it.
However, the part about using the product rule and chain rule still applies though,
at least if he used integration by parts : ).

(btw, d(x(25-x)^3)/dx=(1)(25-x)^3 - 3x(25-x)^2
You might've tried to claim that -(25-x)^2=(x-25)^2, or something like that, but
-3x(25-x)=3x(25-x)(x-25) which is not equal to 3x(25-x)^2)

 

[Sprinal]

you use u substitution

u= 25-x
x=25-u

Damn it... I thougth maybe I could use the U thing... But as I am only in my last year of high school before going to uni doing 3-unit (not the required level to work out what the u is without being told what it is).

I should be the one Slaming my head.