Question

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

Short answer

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

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.

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: \(\int_{a}^{b}[x] \mathrm{d} x = \sum_{k=a}^{b-1} k\)

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: \(\int_{a}^{b} -x dx = -\frac{1}{2}b^2 + \frac{1}{2}a^2\).

Add the two integrals

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

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.