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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?

  • 3
    possible duplicate of [Riema$n$n-Stieltjes Integral](http://math.stackexchange.com/questions/40523/riemann-stieltjes-integral)2011-09-01

0 Answers 0