Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Somebody gives you a function f, but never says what it is. You only know three things: f is continuous, f of x is positive, and f of x times f of minus x equals one. Now find the integral from minus one to one of d x over, one plus f of x. Without knowing f? That sounds impossible, right? By the end of this video, you will do it in four short lines. This chapter is the problem bank for the whole unit, and this puzzle is one of its Part B problems, at JEE Advanced level. So, let's begin!

0:41How to use this chapter

So, let's open this chapter. Part A has seven solved problems, at JEE Main level. Okay? And here is how to use them: attempt Part A as a thirty minute test, after Chapters one to seven; the short dash means to. Right? Now, a quick one. Seven problems, thirty minutes. How much time can you give each problem? Thirty over seven; that is the division sign, and the wavy equals sign means approximately. About four point three, so a little over four minutes a problem. Read it, spot the chapter, and solve. Clear? Then Part B: five longer problems, at JEE Advanced level. Got it? And Part C: four concept checks. They test the ideas that most often come in statement form, true or false. With me?

1:37Part B: two ideas in one problem

Now, for Part B, tell me: what makes a problem longer, at JEE Advanced level? Bigger numbers? Harder integrals? Often, two ideas joined in one question. Each idea alone, you already know from Chapters one to seven. The skill is joining them. Right? For example: the area between y equals x squared and y equals a x. First integrate, to get the area in terms of a, then solve an equation for a. Okay? Next, capital F of x is the integral from zero to x of d t over, one plus t to the power four. The long S is the integral sign, zero and x are its limits, and d t says we integrate in t. Is capital F of x, plus capital F of one over x, constant for x greater than zero? How would you test whether something is constant? Differentiate it! If the derivative is zero for every x greater than zero, it is constant. Capital F of one over x needs Leibniz's rule plus the chain rule, which brings in the derivative of one over x. The little dash after the bracket, called a prime, means derivative, and the derivative of one over x is minus one over x squared. The integrand is continuous and one over x is differentiable for x greater than zero, so the rule applies. Two ideas again. Clear? And the third: an unknown f, with f of x times f of minus x equal to one. Replace x by minus x, then add. That is our puzzle, and our Example two. With me?

3:20Part C: every condition matters

Now, here is one statement from Part C to try. If the integral from a to b of f of x, d x, is greater than or equal to zero, then f is greater than or equal to zero on the closed interval from a to b. The double arrow means implies, and the square brackets include both ends. True or false? It is false, and here is why. Take the integral from minus one to two of x, d x. It is three over two, greater than zero. But x is negative on minus one to zero: the square bracket includes minus one, and the round bracket leaves zero out. Positive integral, yet f is negative on part of the interval. Okay? So, a topper's trick for statements: one counterexample is enough to make a statement false. Right? Now look at this one: the integral from minus one to one of d x over x. One over x is odd, and the limits are symmetric. So by the odd rule, the answer is zero, right? No, no, careful here! One over x is unbounded: its size grows without limit near zero. So this integral does not exist at all. The odd rule needs symmetric limits and an integrable function. Check every condition before you use a rule. Got it?

4:41Example 1: integral of 1/(1 + e to the sin x) from 0 to 2pi

Now, Example one, from Part A, marked easy. Evaluate the integral from zero to two pi of d x over, one plus e to the power sine x. Zero and two pi are the limits, and d x says we integrate in x. Pause the video, and try it first. Done? Okay, let's check. Many students say: from zero to two pi, sine x makes a full turn, and its integral over a full turn is zero. So the answer is zero, right? Oh no, careful, this is the trap! Since e to the power anything is positive, one over, one plus e to the power sine x, is always greater than zero. So capital I, our integral, is greater than zero. It cannot be zero. Okay? Now, replace x by two pi minus x. What does sine x become? Minus sine x, exactly! This is the King property from Chapter four, with a equal to zero and b equal to two pi; the arrow means replace. It keeps the same limits, zero to two pi. Right? So capital I also equals the integral of d x over, one plus e to the power minus sine x. Multiply top and bottom by e to the power sine x. The top becomes e to the power sine x, and the bottom becomes e to the power sine x, plus one. With me? Now add the two forms of capital I together. The tops add to one plus e to the power sine x, which cancels the bottom completely. So two I is the integral of d x from zero to two pi, which is just two pi. Clear? So capital I equals pi. Positive, just as we said. Perfect!

6:30Example 2: 1/(1 + f(x)) from -1 to 1 when f(x)f(-x) = 1

Now, remember our puzzle? Let's finish it. This one is from Part B, marked hard. f is continuous, f of x is greater than zero, and f of x times f of minus x equals one. Evaluate the integral from minus one to one of d x over, one plus f of x. Pause the video, and try it first. Done? Okay, let's check. We do not know f, so we cannot integrate directly. So, which replacement brings in f of minus x? Replace x by minus x, yes! That is the King property again, with a equal to minus one and b equal to one, because a plus b minus x is just minus x. The limits stay minus one to one. Okay? Now, f of x is positive, and f of x times f of minus x is one, so f of minus x equals one over f of x. Then one over, one plus f of minus x, is one over, one plus one over f of x. Multiply top and bottom by f of x, and it becomes f of x over, f of x plus one. The bottom is never zero, because f is positive. Right? Now add the two forms. One over, one plus f of x, plus f of x over, f of x plus one, is just one. Every trace of f is gone! With me? So two I is the integral from minus one to one of d x, which is two. So capital I equals one. Four lines, and we never needed to know f. Puzzle solved!

8:13Quick revision: test plan, replace and add, check conditions

So, quick revision. Part A is a thirty minute test, Part B joins ideas, and Part C tests statements. Replace and add: in the integral from a to b, replace x by a plus b minus x, then add the two forms of capital I together. And check every condition; one counterexample makes a statement false. Now open Part A, set a thirty minute clock, and start solving. All the best!