Modulus, Greatest Integer and Piecewise Integrands

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Draw the sine curve from zero to two pi. One arch above the axis, and one arch below it, right? Now work out the integral from zero to two pi of sine x, d x. You get zero. Zero? So the curve encloses no area at all? Something is off here, isn't it? Today: the modulus, the greatest integer, and other piecewise integrands. Questions like these come up often in JEE, and they all need one habit: split the interval. So, let's begin.

0:37Modulus: split at the zeros

So, rule one: the modulus. The two straight bars around f of x mean mod f of x, okay? We cannot integrate the bars directly, so we split the interval at the zeros of f, the points where f becomes zero. Where f is positive, the bars change nothing. Where f is less than zero, the modulus changes the sign, and the modulus of f becomes minus f. Clear? Back to our puzzle. Between zero and two pi, where is sine x negative? From pi to two pi, yes, exactly! Look at the integral from zero to two pi of sine x, d x. The long S is the integral sign, zero and two pi are the limits, and d x says we integrate in x. The arch from zero to pi gives plus two. The arch from pi to two pi lies below the axis, so it gives minus two. They cancel, and we get zero. That is a signed integral, not the area. With me? Now put the bars on: mod sine x. From pi to two pi, the modulus changes the sign, so that arch also counts as plus two. Two plus two: four. Each arch has area two, so the total is four, not zero. Puzzle solved! And remember, with the lower limit below the upper limit, the integral of a modulus can never come out negative. Got it?

1:52Greatest integer: constant on each piece

Now rule two: the greatest integer function. g of x inside square brackets means the greatest integer less than or equal to g of x, okay? It jumps only when g of x crosses an integer. So we split exactly where g of x crosses an integer. Right? And the reward? On each piece, the greatest integer of g of x is just a constant. Its integral is that constant times the width of the piece, nothing more. Quick question. What is the integral from zero to three of the greatest integer of x, d x? From zero to one, the greatest integer of x is zero. From one to two, it is one. From two to three, it is two. Each piece has width one, so zero plus one plus two: three. Did you get three? Very good!

2:40Fractional part: one period, then repeat

Rule three: the fractional part of x, written as x inside curly brackets. It equals x minus the greatest integer of x. It climbs from zero towards one, drops back to zero, and repeats. So it has period one. Okay? Now tell me: when x is between zero and one, what is the fractional part of x? Just x itself, correct! There, the greatest integer of x is zero. So the integral from zero to one of the fractional part equals the integral of x from zero to one: one half. Clear? It repeats every unit, so each period adds the same one half. That means from zero to n, for a positive integer n, the integral is n over two. Right? For example, from zero to ten: ten periods, ten times one half, that is five. No need to split into ten pieces!

3:37Max and min: split where f=g

And rule four: max and min. Max of f and g, with f and g in curly brackets, takes whichever is bigger at each point; min takes the smaller one. The bigger one can switch over where f equals g, so that is where we split. Okay? Try this. Between zero and one, which is bigger: x, or x squared? x is bigger, yes! Squaring a number between zero and one makes it smaller. They meet at one, and from one to two, x squared is bigger. The sign on screen means greater than or equal to. Right? So the integral from zero to two of max of x and x squared is the integral of x from zero to one, one half, plus the integral of x squared from one to two, seven over three. Total: seventeen over six.

4:25Example 1: mod x and mod of x2 - 1, split at the zeros

Got it? Example one, two quick parts. Evaluate the integral from minus one to two of mod x, d x. Then the integral from zero to two of the modulus of x squared minus one, d x. Pause the video and try it first. Done? Okay, let's check. Mod x is zero at x equals zero, so we split there. From minus one to zero, x is negative, so the modulus of x is minus x. From zero to two, it is just x. Okay? The first piece gives one half, and the second gives two. Total: five over two. Right? Now the second part. Some students just drop the bars and integrate straight from zero to two. Is that allowed? No, no, careful here! x squared minus one is less than zero from zero to one, and greater than zero from one to two. It changes sign at x equals one. Drop the bars, and the negative part cancels some of the positive part: a wrong answer! So split at one. From zero to one, the modulus is one minus x squared, which gives two thirds. From one to two, it is x squared minus one, which gives four thirds. Two thirds plus four thirds: two. That's it!

5:43Example 2: [x2] from 0 to 2 is 5 - root 2 - root 3

Now a harder one. Evaluate the integral from zero to two of the greatest integer of x squared, d x. On the right, the dashed curve is y equals x squared, and the dark steps are y equals the greatest integer of x squared. Use the graph as a hint. Pause the video and try it first. Done? Okay, let's check. The greatest integer jumps at whole numbers. Between zero and two, the only whole number is one. So we split only at x equals one, right? Look at the graph before you answer. Oh no, careful, that is the trap! It is x squared that must cross an integer, not x. As x goes from zero to two, x squared runs from zero to four, so it crosses one, two and three. Okay? x squared equals one at x equals one, two at root two, and three at root three. The tick sign is the square root. See them marked on the graph. With me? So on the four pieces, the greatest integer of x squared is zero, one, two, and three. Those are the step heights in the graph. The first piece adds nothing. Clear? Each step gives height times width. So capital I, our integral, is root two minus one, plus two times, root three minus root two, plus three times, two minus root three. Now collect. The numbers give minus one plus six, that is five. Root two: one minus two leaves minus root two. Root three: two minus three leaves minus root three. Answer: five minus root two minus root three. Beautiful, right?

7:24Quick revision: split at zeros, integers, or where f = g

So, quick revision. Modulus: split at the zeros of f, and change the sign where f is negative. Greatest integer: split where g of x crosses an integer. Fractional part: each period gives one half. And for max and min, split where f equals g. One habit for all of them: split the interval. Now go and try the practice problems on this page. All the best!