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Chapter 2: Problem 13

Show that \(\int_{a}^{b}[x] \mathrm{d} x+\int_{a}^{b}[-x] \mathrm{d} x=a-b\).

Short Answer

Expert verified
Question: Show that the sum of the given definite integrals is equal to a - b: \( \int_{a}^{b}[x] dx + \int_{a}^{b}[-x] dx \) Answer: \(a - b\)

Step by step solution

01

Integrate the greatest integer function [x]

Firstly, let's analyze the greatest integer function [x]. The integral of [x] is the sum of areas of the rectangles formed between the curve y=[x] and the x-axis from a to b, where a and b are integers. To find these areas, we will split the interval into integer subintervals and find the area of each rectangle.
02

Sum the areas of the rectangles

For each integer k in the interval [a, b], the length of the rectangle is k and the width is 1. Thus, the area of each rectangle is given by k*1=k. The definite integral of [x] from a to b is the sum of areas of these rectangles, which can be expressed as: \(\int_{a}^{b}[x] \mathrm{d} x = \sum_{k=a}^{b-1} k\)
03

Integrate the function -x

Now, we will integrate the function -x from a to b. The integral of -x is: \(\int_{a}^{b} -x dx\) . We can find the antiderivative of -x, which is: \(-\frac{1}{2}x^2 + C\). Applying the Fundamental Theorem of Calculus, the definite integral is equal to: \(\int_{a}^{b} -x dx = -\frac{1}{2}b^2 + \frac{1}{2}a^2\).
04

Add the two integrals

Now, we will add our expressions for the two definite integrals: \(\int_{a}^{b}[x] \mathrm{d} x+\int_{a}^{b}[-x] \mathrm{d} x = \sum_{k=a}^{b-1} k -\frac{1}{2}b^2 + \frac{1}{2}a^2\)
05

Simplify the sum and prove the identity

To simplify the sum, we'll apply the formula for the sum of the first n integers: \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\). So, our expression becomes: \(\int_{a}^{b}[x] \mathrm{d} x+\int_{a}^{b}[-x] \mathrm{d} x = \frac{(b-1)b}{2} - \frac{(a-1)a}{2} - \frac{1}{2}b^2 + \frac{1}{2}a^2\) Simplifying, we find that the expression equals: \(a - b\). Thus, we have shown that: \(\int_{a}^{b}[x] \mathrm{d} x+\int_{a}^{b}[-x] \mathrm{d} x=a-b\), as required.

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Most popular questions from this chapter

Show that for each integer \(\mathrm{m}>1\), \(\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{m}<\ln m<1+\frac{1}{2}+\ldots+\frac{1}{m-1}\)

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