Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Integrate one over, one plus e to the power sine x, from zero to two pi. Try substitution, or integration by parts, or partial fractions. You get stuck, right? And yet the answer is a very clean number, in just three lines. How is that possible? This puzzle is from Part A of our unit problem bank: seven problems at the JEE Main level. Problems like these come up often. So, let's begin.

0:33Part A: seven problems, seven ideas

So, here is Part A: seven solved problems, mixed from the whole unit. Problems one and seven are indefinite integrals, so every answer ends with plus capital C, the constant of integration. Okay? Problems two and three are definite integrals. One needs a substitution, and the other needs the King property. Right? Problem four asks for an area. And in problem five, you are given the integral of f, and you must find f itself. Clear? And problem six is the limit of a long sum of fractions. It looks like pure algebra, doesn't it? So which chapter does it really belong to? Integral calculus, yes! It is a sum, but really an integral in disguise. In a mixed set, nobody tells you the chapter. So first name the type, then pick the tool. With me? Today: problems three and six, the easier one first, with one tool for each. Let's go.

1:34Tool 1: the King property

Tool one, from Chapter four: the King property. The long S is the integral sign, a and b are its limits, and d x says we integrate in x. 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, where a and b are the limits of this very integral. Okay? Call this integral capital I, write it both ways, and add. 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 hold the sum. Clear? Now look at the figure: f of x is solid, and f of pi over two minus x is dashed, where pi is the Greek letter. At every point, the two curves add up to one. So in such a case, what is two I? Just b minus a, correct! The integral of one is the length of the interval. So capital I is b minus a, over two. In the figure, each curve takes half the rectangle: pi over four. Right? One more line. For the limits zero and two pi, a plus b minus x is two pi minus x; the arrow means replace x by two pi minus x. And sine of two pi minus x is minus sine x. Got it?

2:59Tool 2: a limit of a sum as an integral

Tool two, from Chapter five: a limit of a sum. The Greek letter sigma means add up, for r from one to n. Lim, with n arrow infinity below it, means the limit as n tends to infinity: n grows without end, and the sideways eight is infinity. So the limit of one over n, times the sum of f of r over n, equals the integral from zero to one of f of x, d x. Okay? Now the figure. The curve here is just a sample, y equals one over, one plus x squared. Each rectangle has width one over n, and height f of r over n. So what does one over n, times the sum, stand for? The total area of the rectangles, exactly! As n tends to infinity, the rectangles get thinner, and that area tends to the area under the curve: the integral. With me? So, the method. Take out one over n, and write every term using only r over n. Right? Then replace r over n by x, and one over n by d x. Clear? And the limits? Careful, not always zero and one! They come from the smallest and largest values that r over n takes, as n tends to infinity. Okay?

4:16Example 1: 1/(1 + e to the sin x) from 0 to 2pi, answer pi

Now, our puzzle from the start: it is problem three of Part A, okay? Evaluate the integral from zero to two pi of d x over, one plus e to the power sine x. Pause the video and try it first. The hint is tool one. Done? Okay, let's check. Let capital I be this integral. Out of habit, many students replace x by pi minus x. Does that help here? No, no, careful here! Sine of pi minus x is just sine x, so nothing changes, and we only get I equals I again. Pi minus x belongs to the limits zero and pi, not to these. Clear? So use this integral's own limits: zero plus two pi, minus x, which is two pi minus x. And sine x becomes minus sine x. Right? So capital I is also the integral of d x over, one plus e to the power minus sine x. Multiply the top and the bottom by e to the power sine x: we get e to the power sine x, over one plus e to the power sine x. Okay? Now add this to the original form. What do the two integrands add up to? One, exactly! Same denominator, and the tops add up to one plus e to the power sine x. With me? So two I is the integral of one from zero to two pi: two pi. The double arrow means therefore: capital I equals pi. Puzzle solved in three lines. Beautiful, isn't it?

5:52Example 2: limit of 1/(n+1) + ... + 1/(3n) is ln 3, not ln 2

Now the harder one, Part A's problem six. Find the limit, as n tends to infinity, of one over n plus one, plus one over n plus two, and so on, up to one over three n; the three dots mean the pattern carries on. Pause the video and try it. The hint is tool two. Done? Okay, let's check. The general term is one over n plus r. The last term is one over three n, so n plus r equals three n. So r runs from one up to two n, not just up to n. Okay? Now take out one over n. One over n plus r equals one over n, times one over, one plus r over n; the dot means times. Every term is now written through r over n. Right? So replace r over n by x, and one over n by d x, to get the integral of d x over one plus x. Limits zero and one, like the formula, so the answer is the natural log of two, right? On screen, l n means the natural log. Oh no, careful, that is the trap! The limits come from the smallest and largest values of r over n. The smallest is one over n, which tends to zero. The largest is two n over n, which is two. So x runs from zero to two. Clear? So we need the integral from zero to two of d x over one plus x. Its antiderivative is log of one plus x. The big square brackets with zero and two mean: put in two, then subtract the value at zero. Log three minus log one, and log one is zero. So the answer is log three, not log two. Very good!

7:39Quick revision: name the type, a + b - x, limits from r/n

So, quick revision. One: in a mixed set, name the type first, and an indefinite integral always gets plus capital C, the constant. Two: the King property replaces x by a plus b minus x, using the integral's own limits. Three: for a limit of a sum, take out one over n, and take the limits from the smallest and largest values of r over n. Now go and try the practice problems on this page. You are ready for them. All the best!