2

Evaluate $$\int_{\frac{2}{3}}^8 f(x)d\alpha(x)$$ where $\alpha$ is continuous and $f$ is the floor function, that is $f(x)$ is the greatest integer less than or equal $x$.

Julien
  • 45,674
Alyse
  • 245
  • 3
  • 9
  • What are your thoughts on this? Maybe you mean $\alpha$ is the floor function? – Pedro Apr 10 '13 at 02:29
  • I know how to evaluate if it is a specific function but an unsure how to when it is in the general sense. It'd make more sense if it were $\alpha$ but the specific problem statement is not. It is $f$ – Alyse Apr 10 '13 at 02:29
  • A related problem. – Mhenni Benghorbal Apr 10 '13 at 02:31
  • I think the problem I am struggling with is that it is $f(x)$ that is the floor function. Do I just treat that simply as $\lfloor x\rfloor$? – Alyse Apr 10 '13 at 02:33
  • The floor function is constant in most places. Split the integral up so that the floor function is constant over the range of integration. – copper.hat Apr 10 '13 at 02:34
  • Do you know how to evaluate $\int_a^b d \alpha(x)$? – copper.hat Apr 10 '13 at 02:44
  • I understand the breaking it up now. I am working to understand the simplification of the integrals to get to the final answer. Are you in agreement that the result should end up being what is shown below in the second answer? – Alyse Apr 10 '13 at 02:45
  • Sue, yes, Marvin's answer is correct. – copper.hat Apr 10 '13 at 04:23

2 Answers2

2

Hint:

Write $\int_{\frac{2}{3}}^8 \lfloor x \rfloor d \alpha(x) = \int_{\frac{2}{3}}^1 \lfloor x \rfloor d \alpha(x)+ \sum_{k=1}^7 \int_{k}^{k+1} \lfloor x \rfloor d \alpha(x) $.

What value does $\lfloor x \rfloor$ take inside these integrals?

copper.hat
  • 178,207
1

$$I = \int_{2/3}^8 \lfloor x\rfloor d \alpha(x) = \int_{2/3}^1 \lfloor x\rfloor d \alpha(x) + \sum_{k=1}^7 \int_k^{k+1} \lfloor x\rfloor d \alpha(x) = 0 + \sum_{k=1}^7 \int_k^{k+1} \lfloor x\rfloor d \alpha(x)$$ Now $\lfloor x \rfloor = k$ for $x \in [k,k+1)$. Hence, we get that \begin{align} I & = \sum_{k=1}^7 k (\alpha(k+1) - \alpha(k))\\ & = \sum_{k=2}^7 \alpha(k) (-k + k-1) - \alpha(1) + 7 \alpha(8)\\ & = 7 \alpha(8) - \sum_{k=1}^7 \alpha(k) \end{align}

EDIT

To make the last step clear, let us explicitly write it out and see. \begin{align} I & = \sum_{k=1}^7 k (\alpha(k+1) - \alpha(k))\\ & = 1 \cdot(\alpha(2) - \alpha(1)) + 2 \cdot(\alpha(3) - \alpha(2)) + 3 \cdot(\alpha(3) - \alpha(2)) + 4 \cdot(\alpha(4) - \alpha(3))\\ & + 5 \cdot(\alpha(5) - \alpha(4)) + 6 \cdot(\alpha(6) - \alpha(5)) + 7 \cdot(\alpha(8) - \alpha(7))\\ & = -\alpha(1) + (1-2) \cdot \alpha(2) + (2-3) \cdot \alpha(3) + (3-4) \cdot \alpha(4) + (4-5) \cdot \alpha(5) + (5-6) \cdot \alpha(6)\\ & + (6-7) \cdot \alpha(7) + 8 \cdot \alpha(8)\\ & = -\alpha(1) - \alpha(2) - \alpha(3) - \alpha(4) - \alpha(5) - \alpha(6) - \alpha(7) + 7 \alpha(8)\\ & = 7 \alpha(8) - \sum_{k=1}^7 \alpha(k) \end{align}