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I think I have understood this concept but wanted to check this question out with you before I plough on using the same method (which may be wrong). I also have a few small questions within the process.

Question:

$$I = \int_0^4 f(x) da(x) $$ where $$f(x) = (x-2)^2$$ and $$a(x) = \left\{\begin{array}{ll} 0 & \text{if }0\leq x \leq 1;\\ \frac{1}{4} & \text{if }1\lt x\lt 3;\\ 1 & \text{if }3\leq x \leq 4. \end{array}\right.$$

Find $L(P)$ and $U(P)$ using a suitable partition and then evaluate I.

Here's my attempt:

$P=\{0,1,2,3,4\}$

$$\begin{align*} L(P) &= [a(1)-a(0)]f(0) + [a(2)-a(1)]f(1) + [a(3)-a(2)]f(2) + [a(4)-a(3)]f(3)\\ &= 0.5f(2) + 0.5f(2) = f(2) = 0.\\ U(P) &= [a(1)-a(0)]f(1) + [a(2)-a(1)]f(1) + [a(3)-a(2)]f(3) + [a(4)-a(3)]f(4)\\ &= 0.5f(1) + 0.5f(3) = 0.5 + 0.5 = 1 \end{align*}$$ so $0\leq I\leq 1$.

Actual Value of I:

Jumps: $0.25f(1) + 0.25f(3) = 0 + 0.25 = 0.25$

Integral: $$\int_1^3 (x-2)^2d(0.25x) = \int_1^3 (x^2-4x+2)0.25\,dx$$ and then we work the integral out and add the jumps to get some number between 0 and 1.

Q1. Is this correct? What do you do if $f(x)$ is just given (i.e. we don't know its formula)?

Q2. Can you not use U(P) and L(P) to work out the integral?

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5  
@user4645: Your answers to my comments on your previous question (math.stackexchange.com/questions/41230/…) display a lack of understanding of how this site works. Please read the FAQ before posting any more questions, and/or explain why you're not accepting any of the answers. Also, please use $\TeX$ to typeset mathematics; this question is very hard to read. –  joriki May 27 '11 at 16:02
    
@user: And you have not accepted nor upvoted any of the answers to the nearly 20 questions you have asked. You should listen to Joriki. –  mixedmath May 27 '11 at 23:06
    
@All: I'm confused. The OP has a temporarily suspended account. Should this question be visible? –  mixedmath May 27 '11 at 23:28
    
@mixedmath: Yes, questions (and answers) posted by suspended users are still visible during the suspension. They cannot interact with them, however. –  Arturo Magidin May 28 '11 at 2:06
3  
possible duplicate of Riemann-Stieltjes Integral –  t.b. Sep 1 '11 at 15:24

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