Main content
Course: IIT JEE > Unit 1
Lesson 1: IIT JEE- Trig challenge problem: arithmetic progression
- IIT JEE perpendicular planes (part 1)
- IIT JEE perpendicular plane (part 2)
- IIT JEE complex root probability (part 1)
- IIT JEE complex root probability (part 2)
- IIT JEE position vectors
- IIT JEE integral limit
- IIT JEE algebraic manipulation
- IIT JEE function maxima
- IIT JEE diameter slope
- IIT JEE hairy trig and algebra (part 1)
- IIT JEE hairy trig and algebra (part 2)
- IIT JEE hairy trig and algebra (part 3)
- Challenging complex numbers problem (1 of 3)
- Challenging complex numbers problem (2 of 3)
- Challenging complex numbers problem (3 of 3)
- IIT JEE differentiability and boundedness
- IIT JEE integral with binomial expansion
- IIT JEE symmetric and skew-symmetric matrices
- IIT JEE trace and determinant
- IIT JEE divisible determinants
- Intersection of circle & hyperbola
- Common tangent of circle & hyperbola (1 of 5)
- Common tangent of circle & hyperbola (2 of 5)
- Common tangent of circle & hyperbola (3 of 5)
- Common tangent of circle & hyperbola (4 of 5)
- Common tangent of circle & hyperbola (5 of 5)
- Trig challenge problem: multiple constraints
- Trig challenge problem: maximum value
- Vector triple product expansion (very optional)
- IIT JEE lagrange's formula
- Representing a line tangent to a hyperbola
- 2010 IIT JEE Paper 1 Problem 50: Hyperbola eccentricity
- Normal vector from plane equation
- Point distance to plane
- Distance between planes
- Challenging complex numbers problem: complex determinant
- Series sum example
- Trig challenge problem: system of equations
- Simple differential equation example
© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice
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.
Want to join the conversation?
- at9:18, is there a proof that d/dx arc-tan x = 1/x^2+1 ? I'm completely unfamiliar with that. Thanks in advance!(6 votes)
- assume arc-tanx is some variable theta. therefore you have 'x=tan(theta)'
differentiate x with respect to theta and you get: {dx/d(theta)} = (sec(theta))^2
taking its reciprocal, we get d(theta)/dx = 1/ (sec(theta))^2
and writing (sec(theta))^2 in terms of x,
(where we know x= tan(theta) and we also know the identity: (sec(theta))^2 = ((tan(theta))^2) +1),
we get (sec(theta))^2= x^2+1
replacing theta with arc-tanx we get:
d(arc-tan x)/dx = 1/x^2+1
I hope that helped you.(20 votes)
- why is the third option not correct, i.e, 0? Because,
22/7 - π = π - π
= 0(5 votes)- 22/7 is not an exact representation of Pi.
Pi is an irrational number i.e., there is no exact representation possible as a ratio of two integers (22 & 7 in the question).
Since the representation is not exact, the difference is not 0.
For some of the possible answers to the question, please refer the link http://www.wolframalpha.com/input/?i=22%2F7+-+pi(7 votes)
- Is it a coincidence that the digits in each row of the Pascals triangle (1:15) are the powers of 11? Like the first row is 11^1 = 1 1, the second is 11^2 = 1 2 1, the third is 11^3 = 1 3 3 1, the fourth is 11^4 = 1 4 6 4 1 etc.(5 votes)
- number 11 hols the secrets of numerology(1 vote)
- Is there a faster way to do this problem
cause in the exam we really can not afford to have 13 mins to solve one question.(1 vote) - pi= 22/7
so, 22/7-pi=22/7-22/7=o
that says option "C" is also correct as in the question it states that they may be a single or multiple answers of the question(1 vote)- 22/7 does not equal pi. It is only an approximation. Therefore, it is erroneous to say that 22/7 - pi = 0.(1 vote)
- At13:36If were getting 22/7-pi. Why wasn't (c) 0 one of the answers?(1 vote)
- Pi is an irrational number, meaning that any fractional or decimal approximation is just that; an approximation. It's a bit strange to think about, but the digits of pi (3.1415926535....) just keep going on and on forever.(1 vote)
- The above explained process is quite lengthy process and why can't you solve it by substituting x with some trigonometric function and also
integral(0 to 1)[f(x)dx]=integral(0 to 1)[f(1-x)dx](1 vote)- Because using that property wont help you. The only way out using elementary functions, is this way, as far as I can see.(1 vote)
- what is improper integration(0 votes)
- If the integrand becomes unbounded at the extreme points of the intervalwhere it is continuous .(1 vote)
- dont criticise ime 10 and i understand this(0 votes)
- Isn't 22/7 the same thing as pie ? so the answer can also be zero right?(0 votes)
- no.. 22/7 is only approximately pi.. they are not the same..(2 votes)
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.