- #1
pc2-brazil
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I was self-studying from a Calculus book and found the following question:
U is the function defined by
U(x)=\begin{cases}<br /> 0 & \text{ if } x<0 \\ <br /> 1 & \text{ if } x\geq 0 <br /> \end{cases}
Let [a, b] be an interval such that a < 0 < b. Show that, even though the function U is discontinuous in [a, b], it is integrable in [a, b] and
\int_a^bU(x)dx=b
Definition of definite integral:
\int_a^bf(x)dx=\lim_{||\Delta||\to 0}\sum_{i=1}^n f(\xi_i)\Delta_ix
where Δ is a partition, Δix is the i-th subinterval of the partition Δ, that is, the subinterval [xi-1, xi], ξi is any number such that xi-1 ≤ ξi ≤ xi and ||Δ|| is the norm of the partition (the largest subinterval of the partition).
I'm not very sure on how to start, and the attempt below seems inconsistent. Could someone point to me what may be wrong with it and suggest corrections?
To show this, I have to show that (according to the definition given by the book I'm using) there is a number L which satisfies the condition that, for any ε > 0, there is a δ > 0 such that
\left | \sum_{i=1}^n U(\xi_i)\Delta_ix \ \right | < \epsilon
for every partition Δ for which ||Δ|| < δ and for any ξi in the closed interval [xi-1, xi], i = 1, 2, ..., n.
Let Δ be a partition of the interval [a,b]. There is a discontinuity at x = 0. So, I will divide the whole sum into two sums: S1 over the interval [a, 0] and S2 over the interval [0, b]. For this, I will specify that x = 0 is a partition point, that is, xk = 0 is at an extreme of a subinterval Δkx.
Calculation of S1: Suppose that in the interval [a, 0] there is no subinterval for which ξi = 0 (that is, for every ξi, U(ξi) = 0). So, the sum of all the U(ξi)Δix in [b, 0) is 0. Thus, S1 = 0.
Now, for S2: Suppose Δjx is the first subinterval for which ξj belongs to [0, b]. Thus, for every ξi in the interval [0, b], U(ξi) = 1; so, the whole Riemann sum will become:
\sum_{i=1}^n f(\xi_i)\Delta_ix = S_1 + S_2 = 0+\sum_{i=j}^n 1\times\Delta_ix=\sum_{i=j}^n \Delta_ix=b
So:
\sum_{i=1}^n f(\xi_i)\Delta_ix - b = 0
and, for every ε > 0, there is a δ > 0 such that
\left | \sum_{i=1}^n f(\xi_i)\Delta_ix - b \right | < \epsilon whenever ||Δ|| < δ
and, thus:
\lim_{||\Delta||\to 0} \sum_{i=1}^n U(\xi_i)\Delta_ix = b
I know this is not complete, because, above, I chose x = 0 to be at an extreme of a subinterval. If x = 0 were in the middle of a subinterval Δkx (that is, if xk-1 ≤ 0 ≤ xk), there would be situations where the Riemann sum would be less than b, and situations in which it would be greater than b, depending on the choice of ξk. But I'm not sure how to handle these situations.
Thank you in advance.
Homework Statement
U is the function defined by
U(x)=\begin{cases}<br /> 0 & \text{ if } x<0 \\ <br /> 1 & \text{ if } x\geq 0 <br /> \end{cases}
Let [a, b] be an interval such that a < 0 < b. Show that, even though the function U is discontinuous in [a, b], it is integrable in [a, b] and
\int_a^bU(x)dx=b
Homework Equations
Definition of definite integral:
\int_a^bf(x)dx=\lim_{||\Delta||\to 0}\sum_{i=1}^n f(\xi_i)\Delta_ix
where Δ is a partition, Δix is the i-th subinterval of the partition Δ, that is, the subinterval [xi-1, xi], ξi is any number such that xi-1 ≤ ξi ≤ xi and ||Δ|| is the norm of the partition (the largest subinterval of the partition).
The Attempt at a Solution
I'm not very sure on how to start, and the attempt below seems inconsistent. Could someone point to me what may be wrong with it and suggest corrections?
To show this, I have to show that (according to the definition given by the book I'm using) there is a number L which satisfies the condition that, for any ε > 0, there is a δ > 0 such that
\left | \sum_{i=1}^n U(\xi_i)\Delta_ix \ \right | < \epsilon
for every partition Δ for which ||Δ|| < δ and for any ξi in the closed interval [xi-1, xi], i = 1, 2, ..., n.
Let Δ be a partition of the interval [a,b]. There is a discontinuity at x = 0. So, I will divide the whole sum into two sums: S1 over the interval [a, 0] and S2 over the interval [0, b]. For this, I will specify that x = 0 is a partition point, that is, xk = 0 is at an extreme of a subinterval Δkx.
Calculation of S1: Suppose that in the interval [a, 0] there is no subinterval for which ξi = 0 (that is, for every ξi, U(ξi) = 0). So, the sum of all the U(ξi)Δix in [b, 0) is 0. Thus, S1 = 0.
Now, for S2: Suppose Δjx is the first subinterval for which ξj belongs to [0, b]. Thus, for every ξi in the interval [0, b], U(ξi) = 1; so, the whole Riemann sum will become:
\sum_{i=1}^n f(\xi_i)\Delta_ix = S_1 + S_2 = 0+\sum_{i=j}^n 1\times\Delta_ix=\sum_{i=j}^n \Delta_ix=b
So:
\sum_{i=1}^n f(\xi_i)\Delta_ix - b = 0
and, for every ε > 0, there is a δ > 0 such that
\left | \sum_{i=1}^n f(\xi_i)\Delta_ix - b \right | < \epsilon whenever ||Δ|| < δ
and, thus:
\lim_{||\Delta||\to 0} \sum_{i=1}^n U(\xi_i)\Delta_ix = b
I know this is not complete, because, above, I chose x = 0 to be at an extreme of a subinterval. If x = 0 were in the middle of a subinterval Δkx (that is, if xk-1 ≤ 0 ≤ xk), there would be situations where the Riemann sum would be less than b, and situations in which it would be greater than b, depending on the choice of ξk. But I'm not sure how to handle these situations.
Thank you in advance.
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