Mixed Problems: Set 1

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take the natural log of tan x, and integrate it from zero to pi over two. Try to find an antiderivative first: substitution, by parts, anything you know. You get stuck, right? Yet the answer comes out in two lines, and it is a very simple number. This is Set one of our mixed problems on integral calculus, and questions like these come up often in JEE Main. So, let's begin.

0:33Set 1: what is inside

So, here is what Set one holds. Nine solved problems, written in the JEE Main style. Each one is short, and each one hides one idea. Okay? They mix the whole unit: standard integrals, the modulus, area, odd functions, Leibniz's rule, limits of sums, and the King property. All mixed up, just like the real exam. Right? Today we take two tools from the earlier chapters, and then two problems from this set: an easy one first, then a harder one. With me?

1:05Tool 1: split the interval

Tool number one, from Chapter three: split the interval. First, the modulus. The two straight bars around f of x are read mod f of x. We split at the zeros of f, the points where f is zero. Where f is less than zero, that is the sign on screen, the modulus changes the sign, so mod f becomes minus f. Okay? Now take the integral of mod f, from a to b. The long S is the integral sign, a and b are its limits, with a below b, and d x says we integrate in x. Tell me: can this integral ever come out negative? Never, correct! Mod f is never negative, so when a is less than b, its integral is greater than or equal to zero; that is the sign on screen. Keep this as a quick check on your answers. Clear? Second, the greatest integer function. g of x inside square brackets means the greatest integer less than or equal to g of x. It jumps only when g of x crosses an integer, so that is exactly where we split. Right? And on each piece, the greatest integer of g of x is a constant. So its integral there is just that constant, times the width of the piece. Got it?

2:18Tool 2: the King property

Tool number two, from Chapter four: the King property. The integral from a to b of capital F of x, d x, equals the integral from a to b of capital F of a plus b minus x, d x. Here a and b must be the limits of this same integral. Okay? Let capital I be this integral. Add the two forms, and two I equals the integral from a to b of F of x, plus F of a plus b minus x, d x. The square brackets just group the sum. Clear? Now think. What if replacing x by a plus b minus x simply turns F into minus F? Yes, exactly! Then the bracket is zero at every point, so two I is zero, and capital I is zero. No antiderivative needed at all. Right? And when the limits are zero and pi over two, with pi the Greek letter, a plus b minus x is pi over two minus x. Then sine of pi over two minus x is cos x, and cos of pi over two minus x is sine x. They simply swap. With me?

3:21Example 1: mod(x - 1) from -2 to 2, [root x] from 0 to 4

Now, the first problem, an easy one: problem number four from the set. Evaluate the integral from minus two to two of mod x minus one, d x. Then the integral from zero to four of the greatest integer of root x, d x. The tick sign is the square root. Pause the video and try it first. Done? Okay, let's check. Part one. Some students just drop the bars and integrate x minus one from minus two to two. Is that allowed? Look what it gives: minus four. The less-than sign says this is below zero. But the integral of a modulus can never be negative, tool one, remember? So dropping the bars is the trap. Okay? So, where is x minus one negative? For x less than one. So we split at x equals one, the zero of x minus one. Right? From minus two to one, the modulus is one minus x, and that piece gives nine over two. From one to two, it is x minus one, which gives one half. Nine over two plus one half: five. Clear? Now part two, the greatest integer of root x. The interval is zero to four, so it jumps at x equals one, two and three, right? No, no, careful here! It jumps only where root x crosses an integer, not where x does. As x goes from zero to four, root x runs from zero to two. It reaches one at x equals one, and two only at x equals four, the very end. Got it? So, just two pieces. The greatest integer of root x is zero from zero up to one, one not included. And it is one from one up to four, four not included. The less-than sign with a line under it means less than or equal to. With me? Each piece gives height times width. Zero times one, plus one times three: three. So the two answers are five and three. Very good!

5:26Example 2: ln tan x and sin 2x ln tan x from 0 to pi/2

Now, remember our puzzle from the start? Let's finish it, with problem number nine from the set. Evaluate the integral from zero to pi over two of the natural log of tan x, d x. On screen, l n means the natural log. And then the same integral with sine two x in front. Pause the video and try it. Hint: use tool two. Done? Okay, let's check. Let capital I be the first integral. The limits are zero and pi over two, so we replace x by pi over two minus x. Sine and cos swap. So what does tan x become? Cot x, exactly! On screen, the first arrow means replace x by pi over two minus x, and the second arrow means becomes. And cot x is one over tan x. Okay? Now the log rule: the natural log of one over tan x is minus the natural log of tan x. So the integrand simply turns into its negative, and capital I equals minus I, right? Bring it across: two I equals zero. The double arrow on screen means implies, so capital I is zero. Puzzle solved: zero, in two lines, with no antiderivative at all. Beautiful, isn't it? Now the second integral, with sine two x in front. Let capital J be this one. Sine and cos swap, so sine two x becomes cos two x, right? Oh no, careful, that is the trap! Put pi over two minus x inside properly. Two times, pi over two minus x, is pi minus two x. And sine of pi minus two x is sine two x again. So sine two x does not change at all. Clear? So only the log flips its sign, and J equals minus J, okay? Two J is zero, so J is zero too. Both answers: zero, and zero. Perfect!

7:20Quick revision: split the interval, King property gives 0

So, quick revision. One: for the modulus and the greatest integer, split the interval first. Two: with the lower limit below the upper, the integral of a modulus is never negative, and the greatest integer of g of x jumps only where g of x crosses an integer. Three: the King property. If replacing x by a plus b minus x turns F into minus F, the integral is zero. Now go and try the practice problems on this page. You can do them. All the best!