MIT integration bee qualifier test

We will do the 2020 MIT Integration Bee qualifier test. This is the test for the people who want to compete in the actual Integration Bee. You would need to know (at least) all the standard integration techniques from calculus 1 and calculu2 such as integration by parts, u sub, trig sub, partial fractions, improper integrals, and more. You may occasionally need some famous results such as the Gaussian Integral, some reduction formulas, symmetric formulas, and more. This test has 20 integrals and you only have 20 minutes. I believe you do not have to show your work and you do not need to put down the +C for indefinite integrals.

0:00 2020 MIT Integration Bee Qualifier Exam
0:22 check out Brilliant
1:20 Q1
2:47 Q2
4:38 Q3
7:36 Q4
10:56 Q5
12:50 Q6
14:07 Q7
17:58 Q8
19:47 Q9
21:10 Q10
22:06 Q11
26:40 Q12
29:59 Q13
32:02 Q14
35:14 Q15
36:20 Q16
37:40 Q17
38:50 Q18
40:05 Q19
40:30 Q20

Equipment:
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[rohanmitra]

Great video! Infact my university held its first integration bee, and im glad to say i won! Did not expect it, but i used a lot of tricks that I saw on this channel before!

[skylardeslypere]

I have an alternate (and faster) solution to Q7.

We know cos²x-sin²x = cos(2x)
Now we also know that sin(x)cos(x) = sin(2x)/2
So we get the integral of 1/16 * sin(2x)^4 cos(2x)

A simple u=sin(2x) will get the job done.

[diablo]

I really like how you use terms like "U world" and "Complex world"
I always find high level math magical, and this really adds to it

[skylardeslypere]

For the first integral, we don't need to worry about x being negative, but that doesn't exclude log(x) from being negative. So the solution should actually be log(2)log(|log(x)|)+log(x).

[thatkindcoder]

I actually learnt quite a bit from this, thx for the video

[sandglass]

For Q11: If you do Integration by parts at the first and then U substitution, that's it. When you do integration by parts it results: -arcsinx/2x² + 1/2 Integral(1/(x²√(1-x²))dx) and the integral can be written as Integral(x^-3[x^-2-1]^-½dx) and with u=x^-2-1 it's done. 😅

[jarkola]

for question 11 you can just use ' by parts ' directly by taking arcsin(x)/x as first function and 1/x^2 as second and it will simplify beautifully.

[TinyMaths]

Believe it or not, your 100 integrals in one video helped me figure out a lot more of the MIT Integration Bee (might have been 2005 or 2006) questions that I could have hoped to answer as I followed MIT's video.
Your examples helped expose a lot of the holes in my knowledge which prepared me for the Bee (I and I can't wait to dig into this video.

[aryirfan]

Sir, you look like you are gradually progressing into becoming the sensei of mathematics!

[bastienhumbert]

Hi! For the 6, we can write: f(x)=sqrt(xsqrt(xsqrt(x...) equivalent to: f(x)=sqrt(x × f(x)) equivalent to: (f(x))^2= xf(x) equivalent to f(x)=x and integrate x

[hassanniaz]

I did Q2 like this:
Let, u= e^x + 1
=> du=e^x dx
=> dx= du/(u-1)
Integral becomes
∫ du/(u*(u-1)) from 2 to infinity
By partial fractions we get,
-ln(u) + ln (u-1) from 2 to infinity
=ln ((u-1)/u)
=ln (1-1/u)
By putting limits we get
=ln (1-0) - ln (1-0.5)
=0-ln (2^-1)
=ln(2)

[alberteinstein]

10:15 if you were to put 1 into that ln argument, you would get ln 0, which is negative infinity. You need to manipulate and then do L’Hôpital’s Rule. Luckily, it would still work out to 0 :)

[mokouf]

First question:
Even if those log are of unknown base (but must be same, since they are both written as log),
result is still the same. log(a) / log(b) = log_b(a).
So we can divide both logs by log(e), making them both becoming ln.

[peterburbery]

I really like practicing integration skills! I would like to see another video on another year's integration bee.

[jimmykitty]

I'm writing eagerly for the Premier ❤

[adityak]

For question 13, using King's rule is better: ∫f(x)dx (a->b) = ∫f(a+b-x)dx (a->b).
I = ∫sin(sinx - x)dx
I = ∫sin(sinx + x)dx
2I = ∫2sin(sinx)cos(x)dx sinx = t
2I = -2cos(sinx) (0 -> 2π)
I = 0

Even for question 15, using King's rule and adding the integrals gives

2I = ∫dx (0->π/2)
I = π/4

The tanx part just cancels out.

[migabok.evariste]

On question 7 you can also use the indentity sin^4(x)•cos^4(x)= 1/16 sin^4(2x)
Then the integral becomes
1/16$ (sin2x)^4 • cos(2x) from here you can make a u-sub. Put u=sin(2x). And finish it off.
😄😆✌👍

[sherlockjunior]

In Q11, what if we directly apply integration by parts,
We get:
(1/2x²)arcsinx +(1/2) ∫ x⁻²dx/√(1-x²)

Here, we have to just calculate the integral:
∫ x⁻²dx/√(1-x²)

If we take x² common from the square root in the denominator, we get:
∫ [x⁻²/x√(x⁻²+1)] dx
--> ∫ [x⁻³/√(x⁻²+1)] dx
Here if we do U substitution of x⁻²+1=t, we get a direct integral of (-1/2)∫dt/√t

I think this is a much faster way than first substitution of x=sin(u) and then applyind DI method! Thanks 😇✌🏻

[xddjqte]

the last question can easily be solved by using gama function because if we just substitute x^4 to y and the do some simple algebra involving calc then we will get integration_0 tp inf_0.25{y^(1/2)e^(-y)}dy which is in gama form. 1/4{gama(1 + 1/2)} = 1/4 * 1/2 * gama(1/2) = 1/4 * pi^(1/2)/2 = pi^(1/2)/8

[holyshit]

7 Maybe double angle would be better
1/16 Int(sin^4(2x)cos(2x)dx) and simple u substitution u = sin(2x)
11 I would calculate it by parts
In first integration by parts I would get rid of arcsinx
then I would rewrite integral
Int(1/(x^2sqrt(1-x^2))dx) as sum of integrals Int(sqrt(1-x^2)/x^2, dx)+Int(1/sqrt(1-x^2), dx)
and integral Int(sqrt(1-x^2)/x^2, dx) again by parts
13. Substitution u = Pi-x and we will get integral of odd function on interval symmetric arount zero
14 If we want to get rid of summation symbol we can use formula fo sum of finite geometric sequence
20. For this rat race this one is quite quick with Gamma function 1/4*Γ(3/2)