Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, a quick puzzle before we start. Take the integral from minus one to one of one over x, d x. One over x is an odd function, and the limits, minus one to one, are symmetric about zero. So the answer is zero, right? Most people say yes in a second. Hold that thought. Today is Part C of our problem bank: four short statements about definite integrals that all sound correct. Is each one really true? Checks like these do appear in JEE, so let's begin.

0:36The odd-function rule, and its fine print

So, first, recall the rule from chapter four. We call f odd when f of minus x equals minus f of x. The long S is the integral sign, minus a and a are its limits, and d x says we integrate in x, okay? For an odd function, the integral from minus a to a of f of x, d x, is zero. Clear? Now our puzzle. Put minus x into one over x, and you get minus one over x. So one over x is odd. The limits are minus one to one, symmetric. So the integral from minus one to one of d x over x is zero, right? No, no, careful here! Look at the point x equals zero. Near zero, one over x becomes as large as you like in size; we say it is unbounded at zero. So this integral does not exist at all. It is not zero, and it is not any other number either. That was the trap in our puzzle. Okay? So the odd-function rule has fine print. The function must be integrable on the closed interval from minus a to a; the square brackets mean the closed interval, with both ends included. Check that first, and only then say zero. Got it?

1:54Odd f: is its integral from 0 to x even?

Next statement. Let f be odd, and let g of x be the integral from zero to x of f of t, d t. Here t is only the variable inside the integral; the top limit, x, is what changes. True or false: g is an even function? Let's test it. Even means g of minus x equals g of x. So write g of minus x: the integral from zero to minus x of f of t, d t. Now put t equals minus u, so d t equals minus d u. With me? But wait. We changed the variable inside a definite integral. What else must change? The limits, yes, exactly! When t is zero, u is zero. And when t is minus x, u is x. Forgetting to change the limits is a very common mistake. Right? So g of minus x is the integral from zero to x of f of minus u, times minus d u. Now f is odd, so f of minus u is minus f of u, and minus times minus is plus. We get the integral from zero to x of f of u, d u. The name of the variable inside does not matter, so this is exactly g of x. Okay? So g of minus x equals g of x, and the statement is true. The double arrow on screen means, so: an odd f gives an even integral from zero to x. Clear?

3:29Example 1: non-negative integral, yet f(x) = x goes negative

Now Example one, an easy one. True or false: if the integral from a to b of f of x, d x, is greater than or equal to zero, then f of x is greater than or equal to zero on the whole interval from a to b. The sign with a line under it means greater than or equal to. Pause the video and try it first. Done? Okay, let's think. The integral is zero or more, so the function must be zero or more too. It sounds so natural, right? Oh no, careful here! The statement is false, and one simple function shows it. Try f of x equals x, on the interval from minus one to two. Okay? The integral of x is x squared over two. The big square brackets, with minus one and two at the side, mean: put in the top limit, then subtract the value at the bottom limit. Two squared over two is two. And the square of minus one is plus one, so at the bottom we get one half. Two minus one half is three over two, which is greater than zero. With me? But on the interval from minus one to zero, x is less than zero, negative. The round bracket at zero means zero itself is left out. So f is negative there, and yet the integral came out positive. Right? So the statement is false. See, to show a statement false, one counterexample is enough. To show one true, like our last statement, we need a proof that works for every f. Perfect!

5:08Example 2: when is the integral of a periodic f periodic?

Now Example two, one step deeper. Let f be periodic with period capital T, so the graph of f repeats after every T units. Is the integral from zero to x of f of t, d t, also periodic? Pause the video and try it first. Done? Okay, first instinct. The function f repeats every period, so its integral should repeat too, right? Not always, and here is the reason. Call the integral capital G of x. And f of x plus T equals f of x for every x; that is what period capital T means. Okay? Now compare G of x plus T with G of x. Their difference is the integral from x to x plus T of f of t, d t. And by the period rule from chapter four, the integral over any one full period is the same as the integral from zero to capital T, over the first period. Right? So each period adds the same fixed amount to capital G, every time. Now tell me, when does G repeat? Exactly when that amount is zero! If the integral from zero to capital T of f is zero, then G of x plus T equals G of x, and G repeats. If it is not zero, every period adds that same amount again and again, so over many periods G drifts further and further away, and never repeats. Clear? Let's check with the course's own pair. First, cos t, which has period two pi. Its integral over one period, from zero to two pi, is zero. And yes, the integral from zero to x of cos t is sine x, which is periodic. Got it? Now one plus cos t, also with period two pi. Over one period its integral is two pi, not zero. So the integral from zero to x is x plus sine x: the x part keeps growing, and it is not periodic. Perfect! So our answer: periodic only when the integral over one period is zero.

7:21Quick revision: 1/x trap, odd to even, sign, periodic test

So, quick revision. One: an odd function from minus a to a gives zero, but only if it is integrable; one over x fails that test. Two: if f is odd, the integral from zero to x is even; and when you substitute, change the limits too. Three: an integral that is zero or more does not force f to be zero or more; one counterexample decides. Four: for f of period capital T, the integral from zero to x is periodic only if the integral over one period is zero. Now try the two practice problems on this page. All the best!