Integration (1 Viewer)

[Pote]

Please help me, it's pretty urgent. Thanks a lot.

 

[香港!]

C)i)
V(x)=(a-x)U(x)+int. U(t) dt {t=0 to t=x}
V'(x)=-U(x)+u(x)(a-x)+d\dx int. U(t) dt {t=0 to t=x}
V'(x)+U(x)=(a-x)u(x)+d\dx int. U(t) dt {t=0 to t=x}
V(x)+int. U(x)dx=int. (a-x)u(x) dx+int. U(t) dt

(a-x)U(x)+int. U(t) dt {t=0 to t=x}+int. U(x) dx=int. (a-x)u(x) dx+int. U(t) dt
(a-x)U(x)+int. U(x) dx=int. (a-x)u(x) dx
int. U(x) dx=int. (a-x)u(x) dx-(a-x)U(x)
int. U(x) dx {from 0 to a}=int. (a-x)u(x) dx {from 0 to a} - (a-x)U(x) {from 0 to a}
int. U(x) dx {from 0 to a}=int. (a-x)u(x) dx {from 0 to a}-[0U(a)-(aU(0))]
int. U(x) dx {from 0 to a}=aU(0)+int. (a-x)u(x) dx {from 0 to a}

 

[香港!]

it's a bit unclear, hope u can understand it :rolleyes

 

[who_loves_maths]

hi Pote,

Part (a):

since the derivative is to be found in terms of 'x', and the integral given is in terms of the variable 't' with one of the ordinates as 'x', then it would appear intuitive to begin by integrating first, then differentiate the integral obtained:

Int[(4t^3 - 3t^2 + t -1)dt] from 'x' to '3' = [t^4 - t^3 + (1/2)t^2 - t] from 'x' to '3'

-----> Int = x^4 - x^3 + (1/2)x^2 - x - (3^4 - 3^3 + (1/2)3^2 -3)

the next step is to differentiate this integral with respect to 'x':

d(Int)/dx = 4x^3 - 3x^2 + x -1 ; note that -(3^4 - 3^3 + (1/2)3^2 -3) of the integral is simply a constant.


Part (b):

same principle as that of Part(a):

Int[(7 - 6t)^4 dt] from '2' to 'x' = [(-1/30)(7 - 6t)^5] from '2' to 'x'

-----> Int = (-1/30)(7 - 12)^5 + (1/30)(7 - 6x)^5

differentiate with respect to 'x':

d(Int)/dx = -(7 - 6x)^4 ; note that (-1/30)(7 - 12)^5 of the integral is simply a constant.


Part (c):

(i)

Differentiating V(x) to get V'(x):

V'(x) = d[(a-x)U(x)]/dx + d[Int[U(t)dt] from 'x' to '0']/dx

Now, d[(a-x)U(x)]/dx can be found using the product rule:

d[(a-x)U(x)]/dx = (a-x).u(x) - U(x)

and, d[Int[U(t)dt] from 'x' to '0']/dx can be found using the approach outlined in Parts (a) & (b):

let W(t) be the primitive of U(t) ;

ie. Int[U(t)dt] from 'x' to '0' = W(x) - W(0) ; but observe that W(0) is a constant.

hence, d[Int[U(t)dt] from 'x' to '0']/dx = d(W(x) - W(0))/dx = U(x)

Therefore: V'(x) = (a-x).u(x) - U(x) + U(x) = (a-x).u(x)


(ii)

in (i) you've shown that V'(x) = (a-x).u(x) ;

hence, Int[V'(x) dx] from 'a' to '0' = Int[(a-x).u(x).dx] from 'a' to '0'

now, LHS = V(a) - V(0) ;
where V(x) = (a-x)U(x) + Int[U(t)dt] from 'x' to '0' = (a-x)U(x) + W(x) - W(0)

ie. LHS = (a-a)U(a) + W(a) - W(0) - (a-0)U(0) - W(0) + W(0) = W(a) - W(0) - a.U(0)

but since Int[U(x)dx] = W(x) ; then W(a) - W(0) = Int[U(x)dx] from 'a' to '0'

ie. LHS = (Int[U(x)dx] from 'a' to '0') - a.U(0)

Therefore, substitute this expression for the LHS back:

(Int[U(x)dx] from 'a' to '0') - a.U(0) = Int[(a-x).u(x).dx] from 'a' to '0'

-----> Int[U(x)dx] from 'a' to '0' = a.U(0) + Int[(a-x).u(x).dx] from 'a' to '0'

as required.


hope that helps [i hope you understand what i typed, some of the notation might be confusing for a while, so sorry about that.]


Edit: FinalFantasy beat me to it as per usual, though i was typing out for all the parts of that question you posted, didn't realise you only needed help for Part (c)...

 

[acmilan]

You guys beat me to it, but heres what I was typing anyways:

This question requires knowledge of the Fundamental Theorem of Calculus. Using this theorem it can be shown that:

where c is a constant.

The first question looks exactly like the above, thus the answer is x³ - 3x² + x - 1.

The second question is almost in the above form, except the x and 2 need to switch positions in the integral. This can be done by multiplying the integral by -1. Hence the solution is -(7 – 6x)⁴.

For part (c):

 

[香港!]

ahh, juz read AC and WLM's solutions and learnt something!
i didn't know that:
d\dx int. f(t) dt (from 0 to x)=f(x) -_-

 

[acmilan]

Not necessarily from 0 to x, its from c to x, where c can be any constant.

 

[香港!]

@_@
123456789

 

[Pote]

thank you! The explanations are very understandable too!