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Course: IIT JEE > Unit 1

Lesson 1: IIT JEE

IIT JEE integral with binomial expansion

2010 IIT JEE Paper 1 Problem 41 Integral with Binomial Expansion and Algebraic Long Division. Created by Sal Khan.

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Video transcript

The values of the definite integral from 0 to 1 of x to the fourth times 1 minus x to the fourth, all of that over 1 plus x squared dx is or are-- and they say are because more than one of these might be the correct answer. This is one of those multiple correct answer problems. So this is just a straight-up definite integral. And it looks like the best way to do it is really just expand out the numerator, see if we can simplify that with the denominator, and then just try to take the antiderivative. So let's do that. So first, let's expand 1 minus x to the fourth. And when you're taking a binomial to the fourth power, to remember the coefficients, you, of course, can use the traditional binomial coefficients. But I find it easier just to use Pascal's triangle to remember what the coefficients are. So if you have a binomial, if you square it, the coefficients are 1, 1 plus 1 is 2, and then 1. And I've done several videos where I explain why this works. And if you take it to the third power, the coefficients are 1, 3, 3, and 1. I'm just adding the two branches that point to that one node there. I guess you could view it that way. And then if you take it to the fourth power, the coefficients are 1, 4, 6, 4, and 1. So if we want to take 1 minus x to the fourth power-- and I'm going to write it like this. I'm going to write it as negative x plus 1 to the fourth power. And I want to do it that way because I want the high-degree x terms to be listed first. Then this is going to be-- so you're going to start with this term at the highest power, then this term at the 0 power. And then each successive term, this term is going to lower in power, and this one's going to go up in power. So it's going to be negative x to the 1/4 times 1 to the 0-- I'll just write it here-- times 1 to the 0 plus-- the coefficient of that first one was just that 1, so plus 4 times negative x to the third times 1 to the 1 power, plus-- now we have our 6-- plus 6 times negative x squared times 1 squared. Or another way to think about is the exponents on the 1 and the negative x should always add up to 4. And then you have plus-- now we're at this 4-- plus 4 times negative x times 1 to the third, and then plus 1 to the fourth power, or negative x to the 0 times 1 to the fourth, which is really just 1 to the fourth power. So this is going to simplify to-- so negative x to the fourth. Negative x to the fourth is the same thing as x to the fourth. Negative 1 to the fourth is just 1 times 1 plus-- well, actually, I should say minus, because negative x to the third is negative x to the third, so this is minus 4x to the third. So we've done that term. This term will become positive. Negative x squared is the same thing as x squared. So it's plus 6x squared. And then this is negative 4x plus 1. Now, we want to multiply this whole thing that we just expanded times x to the fourth. So let's multiply that whole thing times x to the fourth. So the expression x to the fourth times 1 minus x, all of that to the fourth power is going to be equal to-- we'll just multiply this times x to the fourth. That's pretty straightforward. It's going to be x to the eighth minus 4x to the seventh plus 6x to the sixth minus 4x to the fifth plus x to the fourth. Now we want to divide this bottom expression into the top expression. So we can just use some algebraic long division. So we want to divide-- and I'll write it as x squared plus 1. We want to divide x squared plus 1 into this big thing, into this thing right here. So let me copy and paste that. We want to divide it into this, and we're going to break out some algebraic long division to do it. Let me draw the division symbol right over there. And so we just say we'd look at the highest-degree terms on every step. x squared goes into x to the eighth x to the sixth times. So we put in the x to the sixth place. x to the sixth times x squared is x to the eighth. x to the sixth times 1 is x to the sixth. And then we subtract this from the number up here, the expression up here. x to the eighth minus x to the eighth is nothing. Negative 4x to the seventh minus nothing is negative 4x to the seventh. 6x to the 6 minus x to the sixth is plus 5x to the sixth. And then these, you're subtracting nothing from them, so it's minus 4x to the 5th plus x to the fourth. Now, x squared goes into x to the seventh, or it goes into negative 4x to the seventh, negative 4x to the fifth times. Let's scroll down a little bit. Negative 4x to the fifth times x squared is negative 4x to the seventh. Negative 4x to the fifth times 1 is negative 4x to the fifth. Now we subtract this from that. These cancel out. These cancel out. And we are left with 5x to the sixth plus x to the fourth. x squared goes into 5x to the sixth 5x to the fourth times, so plus 5x to the fourth. I just arbitrarily switched colors since yellow was becoming monotonous. 5x to the fourth times x squared-- scroll down some more. 5x to the fourth times x squared is 5x to the sixth. 5x to the fourth times 1 is 5x to the fourth. And now we subtract. We just subtract. Those cancel out. x to the fourth minus 5x to the fourth is negative 4x to the fourth. Nothing to bring down. x squared goes into negative x to the fourth. It goes into it negative 4x squared times. Negative 4x squared times x squared is negative 4x to the fourth. Negative 4x squared times 1 is negative 4x squared. And now we subtract that from that. You subtract this from 0, where 0 minus negative 4x squared is going to be positive 4x squared. All of these steps, we're always subtracting. Maybe I should draw the subtraction symbol right over there. And then x squared goes into 4x squared. We're in the home stretch. It goes into it 4 times, so plus 4. 4 times x squared is 4x squared. 4 times 1 is 4. And then we subtract one last time. We subtract this whole thing. These cancel out. And then we're just left with a remainder of negative 4. So this whole thing, this whole expression over here, simplifies to this plus negative 4 over x squared plus 1. Let me rewrite it. Let me copy and paste it. So it turns into this. So copy and then let me paste it down here. I wasted a lot of real estate. Paste. So it turns into this, this, and then we have the minus 4 over the x squared plus 1. So minus 4 over x squared minus 4 over x squared plus 1. So this is the entire expression that we have inside of the integral. So this is the same thing as the integral of this from 0 to 1 times dx. So now we just have to take the antiderivative of this. And this looks pretty straightforward. The only complicated part is this right here. But you might recognize that the derivative with respect to x of arctangent of x is equal to 1 over x squared plus 1. So that kind of simplifies this part for us. So let's tackle the problem. So let's just take the antiderivative. This is pretty straightforward. Antiderivative, we just have to make sure we don't make a careless mistake. Antiderivative of x to the sixth is going to be x to the seventh over 7. Or maybe I should say 1/7 x to the seventh. Antiderivative of this is negative 4/6, which is the same thing as negative 2/3 x to the sixth. Antiderivative of that is x to the 5th, plus x to the fifth, right? I'm just raising the exponent. So that's where I get the x to the fifth. And then I divide by that exponent. So 5 divided by 5 is 1. And then I have x to the third, or it's negative 4/3 x to the third plus 4x, and then minus 4 arctangent of x. So let's think about this a little bit. So we're going to evaluate it at 1 and then subtract from that it, and evaluate at 0. So we need to evaluate this at 1, and then from that, subtract it at 0. So if you evaluate it at 1, you have 1/7 minus 2/3 plus 1 minus 4/3 plus 4. That's plus 4 right over there, and then minus 4 arctangent of 1. So let's think about what the arctangent of 1 is. So the arctangent of 1 is essentially saying if I have a right triangle, if I take the opposite over the adjacent-- so the opposite is telling us give us the angle. So we're looking for this angle. We're looking for the angle where the tangent of this angle is equal to 1. Or another way to say it is that the opposite over the adjacent is equal to 1. Or another way to say it is the opposite is equal to the adjacent. So this side has to be equal to this side. This is going to be a slope of 1 right over here. Now, we could try to look at different types of triangles. And it might pop out at you what kind this is, but you could just go to basic geometry to think about this. This is a right angle right here. These two sides are equal. These two angles are equal. So these two angles have to evenly split the remaining 90 degrees in the triangle because it adds up to 180. So this is a 45-degree angle. But in general, when they don't tell you whether you're in radians or degrees, at least on most exams I've seen, you should just assume that you are in radians. So 45 degrees is the same thing-- let's see. Pi is 180 degrees. Pi/2 is 90 degrees. Pi/4 is 45 degrees. So this is equal to-- that right over there is pi/4 radians. So this over here is pi/4 radians, or I could just say pi/4. Or another way to think about it, tangent of pi/4 is equal to 1. So this is just the inverse function. So these two cancel out. So I just evaluate it at 1. Before I move forward, let me subtract from that this whole thing evaluated at 0. And this is a little bit more straightforward. All of these things are going to be 0 when you evaluate it at 0. And what's the arctangent of 0? Well, tangent of 0 is 0. If this angle is 0, if you just had a flat line, then your slope is flat. So this is just going to be 0. So this whole thing evaluated at 0 is going to be 0. So the whole value of our definite integral is going to be this thing over here. So it's going to be all of this stuff, and then this last term right over here is negative 4 times pi/4, so it's going to be negative pi. And all of these are going to add up to some number without a pi in it. So if you are under time pressure, it's actually a good idea to just look at the answers before we even add up all of those fractions and say, which of these have a negative pi in them? And only one of them has a negative pi in them, so our answer is most probably going to be this. If you're under time pressure, I would just circle that. It's clearly not any of these. None of these have a negative pi in them. And there's no way that that negative pi can get altered by this stuff over here. So you wouldn't have to worry about adding it. But since we want to make sure we have the right answer, let's add up all of these fractions. So let's think about what we can do here. So let's see. Negative 2/3 minus 4/3, let me take those two terms. Negative 2/3 minus 4/3 is negative 6/3, or negative 2. So those become negative 2 plus 1 plus 4, and then plus 1/7. So this is 3 plus 1/7. And of course, we have the negative pi out there, minus pi. And this is the same thing as 21/7 plus 1/7 minus pi, which is equal to 22/7 minus pi. So that is indeed the only answer-- 22/7 minus pi. And we're done.