- #1

- 205

- 3

I was self-studying from a Calculus book and found the following question:

U is the function defined by

[tex]U(x)=\begin{cases}

0 & \text{ if } x<0 \\

1 & \text{ if } x\geq 0

\end{cases}[/tex]

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

[tex]\int_a^bU(x)dx=b[/tex]

Definition of definite integral:

[tex]\int_a^bf(x)dx=\lim_{||\Delta||\to 0}\sum_{i=1}^n f(\xi_i)\Delta_ix[/tex]

where Δ is a partition, Δ

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

[tex]\left | \sum_{i=1}^n U(\xi_i)\Delta_ix \ \right | < \epsilon[/tex]

for every partition Δ for which ||Δ|| < δ and for any ξ

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: S

Calculation of S

Now, for S

[tex]\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[/tex]

So:

[tex]\sum_{i=1}^n f(\xi_i)\Delta_ix - b = 0[/tex]

and, for every ε > 0, there is a δ > 0 such that

[tex]\left | \sum_{i=1}^n f(\xi_i)\Delta_ix - b \right | < \epsilon[/tex] whenever ||Δ|| < δ

and, thus:

[tex]\lim_{||\Delta||\to 0} \sum_{i=1}^n U(\xi_i)\Delta_ix = b[/tex]

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 Δ

Thank you in advance.

## Homework Statement

U is the function defined by

[tex]U(x)=\begin{cases}

0 & \text{ if } x<0 \\

1 & \text{ if } x\geq 0

\end{cases}[/tex]

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

[tex]\int_a^bU(x)dx=b[/tex]

## Homework Equations

Definition of definite integral:

[tex]\int_a^bf(x)dx=\lim_{||\Delta||\to 0}\sum_{i=1}^n f(\xi_i)\Delta_ix[/tex]

where Δ is a partition, Δ

_{i}x is the i-th subinterval of the partition Δ, that is, the subinterval [x_{i-1}, x_{i}], ξ_{i}is any number such that x_{i-1}≤ ξ_{i}≤ x_{i}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

[tex]\left | \sum_{i=1}^n U(\xi_i)\Delta_ix \ \right | < \epsilon[/tex]

for every partition Δ for which ||Δ|| < δ and for any ξ

_{i}in the closed interval [x_{i-1}, x_{i}], 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: S

_{1}over the interval [a, 0] and S_{2}over the interval [0, b]. For this, I will specify that x = 0 is a partition point, that is, x_{k}= 0 is at an extreme of a subinterval Δ_{k}x.Calculation of S

_{1}: Suppose that in the interval [a, 0] there is no subinterval for which ξ_{i}= 0 (that is, for every ξ_{i}, U(ξ_{i}) = 0). So, the sum of all the U(ξ_{i})Δ_{i}x in [b, 0) is 0. Thus, S_{1}= 0.Now, for S

_{2}: Suppose Δ_{j}x is the first subinterval for which ξ_{j}belongs to [0, b]. Thus, for every ξ_{i}in the interval [0, b], U(ξ_{i}) = 1; so, the whole Riemann sum will become:[tex]\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[/tex]

So:

[tex]\sum_{i=1}^n f(\xi_i)\Delta_ix - b = 0[/tex]

and, for every ε > 0, there is a δ > 0 such that

[tex]\left | \sum_{i=1}^n f(\xi_i)\Delta_ix - b \right | < \epsilon[/tex] whenever ||Δ|| < δ

and, thus:

[tex]\lim_{||\Delta||\to 0} \sum_{i=1}^n U(\xi_i)\Delta_ix = b[/tex]

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 Δ

_{k}x (that is, if x_{k-1}≤ 0 ≤ x_{k}), 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 ξ_{k}. But I'm not sure how to handle these situations.Thank you in advance.

Last edited: