Substitution and Integration by Parts in Definite Integrals

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Integrate x times the square root of one minus x squared, from zero to one. Between zero and one, this function is never negative, so the answer must be positive, right? But a student substitutes t equals one minus x squared, and gets minus one third. Something has gone wrong; today we find exactly where. Our topic: substitution and integration by parts in definite integrals. Questions on these come up often in JEE, so let's begin.

0:37Substitution: the limits change too

So, substitution with limits. First, the conditions. f is continuous on the closed interval a to b; the square brackets mean the end points are included. And x equals g of t, where g is continuously differentiable: its derivative exists and is continuous. Okay? Now think. x runs from a to b. When we change over to t, where does t run? Yes, exactly! Between the values of t that give x equal to a, and x equal to b. We call them alpha and beta, the Greek letters on screen. So, g of alpha equals a, and g of beta equals b. Right? Now the result. The long S is the integral sign, with the lower limit a at the bottom and the upper limit b at the top. The integral from a to b of f of x, d x, equals the integral from alpha to beta of f of g of t, times g prime of t, d t. g prime, with the little dash, is the derivative of g. Clear? Now, once we integrate in t, do we go back to x before putting in the limits? No need at all! Alpha and beta already belong to t. So the rule is: change the limits with the variable, and never return to x. Mixing a t answer with the old x limits is exactly the mistake in our puzzle. Got it?

2:05Integration by parts with limits

Now, by parts with limits. u is the factor we differentiate; v prime, with the dash, is the factor we integrate, so v is its integral. The integral from a to b of u v prime, d x, equals u v in square brackets from a to b, minus the integral from a to b of u prime v, d x. With me? See the small a and b on that square bracket. What do they mean? Correct: put in the upper limit, then subtract the lower limit. u of b times v of b, minus u of a times v of a. So the limits go on both parts, and the bracket is just a number, so we work it out straight away. Okay?

2:49Example 1: x root(1 - x2) and x/(1 + x4) from 0 to 1

Now, our puzzle, as Example one. Evaluate the integral from zero to one of x times the square root of one minus x squared, d x; the tick-shaped sign is the square root. And the integral from zero to one of x over one plus x to the power four, d x. Pause the video and try them first. Done? Okay, let's check. Put t equals one minus x squared. Then d t equals minus two x d x, so x d x equals minus one half d t. Right? Now, our puzzle student kept the old limits, zero to one, and got minus one third. Is that allowed? No, no, careful here! Zero and one are limits for x, not for t. The double arrow means gives: x equals zero gives t equals one, and x equals one gives t equals zero. Okay? So we get minus one half, times the integral from one to zero of root t, d t. Swapping the limits changes the sign, so the minus goes away: one half times the integral from zero to one. With me? The integral of root t is two thirds of t to the power three over two; the dot means times. From zero to one, the bracket is one. So one half times two thirds: one third, positive! Puzzle solved: that minus came from the old limits. Now the second one: x on top, and x to the power four is x squared, whole squared. So which substitution? Yes, t equals x squared! Then d t equals two x d x. x equals zero gives t equals zero, and x equals one gives t equals one. Same limits this time, but we still check them. Clear? So we get one half, times the integral from zero to one of d t over one plus t squared. That is one half of tan inverse t, from zero to one; the minus one up top means inverse, not one over. Tan inverse one is pi over four, with pi the Greek letter, and tan inverse zero is zero. So the answer is pi over eight. Very good!

5:05Example 2: (ln x)2 from 1 to e and e^x sin x from 0 to pi/2

Example two, with integration by parts. Evaluate the integral from one to e of natural log of x, whole squared, d x; l n is the natural log, and e is its base, the number e. And the integral from zero to pi over two of e to the power x, times sine x, d x. Pause the video and try them first. Done? Okay, let's check. The first one has only one function: log x, whole squared. So for by parts, what do we take as v prime? Yes, exactly, v prime equals one, so v is x! And u is log x, whole squared. The bracket is x times log x whole squared, from one to e, minus the integral of x times u prime. And u prime, by the chain rule, is two log x over x. Okay? Now the bracket, straight away. At e, log e is one, so we get e; at one, log one is zero, so we get zero. In the integral, the x cancels, leaving two log x. And the integral of log x is x log x minus x, from the by parts lesson. Got it? Now a quick trap. At x equals one, log one is zero. So is this whole bracket zero there? No, it is not, so be careful here! x log x is zero there, but minus x is minus one. So the bracket is e minus e, which is zero, minus the value minus one: that gives one. And the answer is e minus two times one: e minus two. Right? Now e to the power x times sine x, which we met in the by parts lesson. Its integral is one half e to the power x, times sine x minus cos x, plus capital C, the constant of integration. In a definite integral, that constant cancels when we subtract. With me? Now the limits, zero to pi over two. Tell me first: at x equals zero, what is sine x minus cos x? It is zero minus one, which is minus one. At pi over two, sine is one and cos is zero, so the top value is one half e to the power pi over two. At zero, e to the power zero is one, so the bottom value is one half times minus one. Clear? Subtract, and watch the signs: minus one half, times minus one, gives plus one half. So the answer is one half, times e to the power pi over two, plus one. Lose that minus sign, and the answer goes wrong.

7:40Quick revision: change limits, [uv] from a to b, sign check

So, quick revision. Substitution: change the limits with the variable, and never return to x. By parts: u v in brackets from a to b, minus the integral of u prime v; the limits go on both parts. Swap the limits, change the sign. And if f is greater than or equal to zero on a to b, with a less than b, the answer is also greater than or equal to zero. That check caught our puzzle! Now go and try the practice problems on this page. All the best!