Prove f is integrable using six subintervals and evaluate.

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In summary, f is defined piecewise as 2 for 0≤x<1, 0 for x=1, -1 for 1<x<2, 3 for x=2, 0 for 2<x<3, and 1 for x=3. To prove that f is integrable, six subintervals are used (0, 1-h), (1-h, 1), (1, 1+h), (1+h, 2-h), (2-h, 2), (2, 3). The discontinuities at 2 and 3 are ignored since they do not contribute to the integral. Using the partition P and the Riemann sum, the
  • #1
gotmilk04
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Homework Statement


Define f as: f(x)= 2 if 0[tex]\leq[/tex]x<1
f(1)=0
f(x)= -1 if 1<x<2
f(2)= 3
f(x)=0 if 2<x<3
f(3)=1
Prove f is integrable using six subintervals and find the value of [tex]\int[/tex]f(x) dx


The Attempt at a Solution


Let P={0, 1-h, 1+h, 2-h, 2+h, 3-h, 3} where 0<h<1/2
I just need help proving [tex]\int[/tex]f(x) dx=1.
I can prove I[tex]\geq[/tex]1, but I'm having trouble proving J<1+[tex]\epsilon[/tex]
I have S[tex]_{P}[/tex]=2(1-h)+2(2h)+(-1)(1-2h)+3(2h)+0(1-2h)+h= 2+9h
but now I don't know how to get that to less than 1+[tex]\epsilon[/tex]
 
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  • #2
gotmilk04 said:

Homework Statement


Define f as: f(x)= 2 if 0[tex]\leq[/tex]x<1
f(1)=0
f(x)= -1 if 1<x<2
f(2)= 3
f(x)=0 if 2<x<3
f(3)=1
Prove f is integrable using six subintervals and find the value of [tex]\int[/tex]f(x) dx


The Attempt at a Solution


Let P={0, 1-h, 1+h, 2-h, 2+h, 3-h, 3} where 0<h<1/2
I would split the interval up this way:
P = {0, 1 - h, 1, 1 + h, 2 - h, 2, 3}

I'm basically ignoring the discontinuities at 2 and 3, since they aren't going to contribute anything to the integral.
gotmilk04 said:
I just need help proving [tex]\int[/tex]f(x) dx=1.
I can prove I[tex]\geq[/tex]1, but I'm having trouble proving J<1+[tex]\epsilon[/tex]
I have S[tex]_{P}[/tex]=2(1-h)+2(2h)+(-1)(1-2h)+3(2h)+0(1-2h)+h= 2+9h
but now I don't know how to get that to less than 1+[tex]\epsilon[/tex]
 

1. How do you prove that f is integrable?

To prove that f is integrable, you can use the Riemann sum method. This involves dividing the interval of integration into smaller subintervals and calculating the sum of the areas of the rectangles formed by the function and the subintervals. If the limit of these sums exists as the subintervals become smaller, then f is integrable.

2. How many subintervals are needed to prove integrability?

In general, at least two subintervals are needed to prove integrability. However, for a more accurate evaluation of the integral, it is recommended to use at least six subintervals.

3. How are the subintervals chosen?

The subintervals are usually chosen to be of equal width, but it is not a requirement. The endpoints of the subintervals are chosen based on the limits of integration and the desired number of subintervals.

4. What is the significance of using six subintervals?

Six subintervals provide a good balance between accuracy and efficiency. With six subintervals, the Riemann sum method can give a reasonably accurate evaluation of the integral without requiring too much computation.

5. Can the number of subintervals be increased for a more accurate evaluation?

Yes, increasing the number of subintervals can lead to a more accurate evaluation of the integral. However, this also increases the amount of computation required. It is important to strike a balance between accuracy and efficiency when choosing the number of subintervals.

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