2011 MIT Integration Bee Qualifying Test

**Random Variable** · January 26th 2011, 02:19 AM

http://web.mit.edu/abhinavk/www/inte...e/qual2011.pdf

Nothing terribly exciting. But you're only given 20 minutes to answer 25 questions. Feel free to post a solution or two.

Moderator edit: Moved to this subforum from Calculus (because I think the intent of the post is to provide a challenge ....?)

**TheCoffeeMachine** · January 26th 2011, 02:41 AM

Originally Posted by Random Variable

...20 minutes...

That's ridiculous, though, isn't it? The first one was meant to be inviting:

$\begin{aligned} \displaystyle ~ 1. ~ \int \frac{x^6-1}{x^4+x^3-x-1}\;{dx} & = \int \frac{(x^2-1)(x^2-x+1)(x^2+x+1)}{(x^2-1)(x^2+x+1)}\;{dx} \\&= \int x^2-x+1 \;{dx}= \frac{1}{3}x^3-\frac{1}{2}x+x+k. \end{aligned}$

I like the third one -- if you tweak it, the answer is obvious:

$\displaystyle 3. \int \frac{2x}{\sqrt{1-x^4}}\;{dx} = \int \frac{(x^2)'}{\sqrt{1-(x^2)^2}}\;{dx} = \sin^{-1}\left(x^2\right)+k.$

The same goes for the 19th:

$\displaystyle 19. \int \frac{4x}{1-x^4}\;{dx} = 2\int \frac{(x^2)'}{1-(x^2)^2}\;{dx} = 2\tanh^{-1}\left(x^2\right)+k.$

For the 12th, the quickest way I know of is:

$\displaystyle \begin{aligned} 12. & \int \frac{1}{\cos{x}}\;{dx} = \int\frac{\cos{x}}{1-\sin^2{x}}\;{dx} = \int\frac{\cos{x}}{(1+\sin{x})(1-\sin{x})}\;{dx} \\& = <br /> \frac{1}{2} \int\frac{\cos{x}}{1+\sin x}\;{dx}-\frac{1}{2} \int\frac{-\cos{x}}{1-\sin x}\;{dx} = \frac{1}{2}\ln\bigg|\frac{1+\sin{x}}{1-\sin{x}}\bigg|+k.\end{aligned}$

**Random Variable** · January 26th 2011, 07:33 AM

That's ridiculous, though, isn't it?

If you have to stop to think, you're in trouble.

2) $\displaystyle \int (2 \ln x + \ln^{2}(x)) \ dx = \int (x \ln^{2})' \ dx = x \ln^{2}(x) + C$

4) $\displaystyle \int \frac{x^{2}+1}{x+1} \ dx = \int \frac{x^{2}}{x+1} \ dx + \int \frac{dx}{x+1} = \int x \ dx - \int \frac{x}{x+1} + \int \frac{dx}{x+1}$

$= \displaystyle \int x \ dx - \int dx + \int \frac{dx}{x+1} + \int \frac{dx}{x+1} = \frac{x^{2}}{2} - x + 2 \ln |x+1| + C$

**Chris L T521** · January 26th 2011, 08:13 AM

Here's another way to do 4:

$\begin{aligned}<br /> \int\frac{x^2+1}{x+1}\,dx &= \int (x+1)\,dx - \int\frac{2x}{x+1}\,dx\\ &= \int(x-1)\,dx+ 2\int\frac{\,dx}{x+1}\\ &= \tfrac{1}{2}x^2-x+2\ln|x+1|+C\end{aligned}$

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Here's 8:

$\begin{aligned}<br /> \int\sqrt{\csc x-\sin x}\,dx &= \int\frac{\cos x}{\sqrt{\sin x}}\,dx\\ &=\int\frac{(\sin x)^{\prime}}{\sqrt{\sin x}}\,dx \\ &= 2\sqrt{\sin x}+C\end{aligned}$

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Here's another way for 12 (note 1/cos x = sec x):

$\begin{aligned}<br /> \int\sec x\,dx &= \int\frac{\sec^2x+\sec x\tan x}{\sec x+\tan x}\,dx\\ &=\int\frac{(\sec x+\tan x)^{\prime}}{\sec x + \tan x}\,dx\\ &= \ln|\sec x+\tan x|+C\end{aligned}$

Rmk: it involves some algebra, but TCM's solution is the same as mine.

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16:

$\begin{aligned}\int\frac{1}{\log x}+\log(\log x)\,dx &= \int\left(x\log(\log x)\right)^{\prime}\,dx \\ &= x\log(\log x)+C\end{aligned}$

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20:

$\begin{aligned}\int x^x(1+\log x)\,dx &= \int (x^x)^{\prime}\,dx\\ &= x^x+C\end{aligned}$

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21:

$\displaystyle\int_0^6\sqrt{6x-x^2}\,dx = \tfrac{9}{2}\pi$ by geometry (this is the area of the upper half-circle of radius 3).

**FernandoRevilla** · January 26th 2011, 08:50 AM

$5.\;\;\displaystyle\int\dfrac{\sin^3x+\sin^2x-2\sin x-2}{\sin^2x+2\sin x+1}dx=$

$\displaystyle\int\dfrac{(\sin x+1)(\sin^2x-2)}{(\sin x+1)^2}dx=$

$\displaystyle\int\left(\sin x-1-\dfrac{1}{\sin x+1}\right)dx=\ldots$

P.S. What on earth is the philosophy of that test?

Fernando Revilla

**Random Variable** · January 26th 2011, 11:41 AM

6) Let's ignore that fact that it's a common integral.

$\displaystyle \int \frac{1}{\sinh^{2}x} \ dx = \int \frac{4}{(e^{x}-e^{-x})^{2}} \ dx$

$\displaystyle = 4 \int \frac{e^{2x}}{(e^{2x}-1)^{2}} \ dx = 2 \int \frac{1}{u^{2}} \ du = -\frac{2}{u} + C$

$\displaystyle = - \frac{2}{e^{2x}-1} + C = 1 - \frac{e^{2x}-1}{e^{2x}+1} + C$

$= -\coth(x) + C_{2}$

**Random Variable** · January 26th 2011, 11:57 AM

23) $\displaystyle \int x e^{e^{x^{2}}+x^{2}} \ dx$

$\displaystyle = \frac{1}{2} \int e^{e^{u}+u} \ du = \frac{1}{2} \int e^{e^{u}}e^{u} \ du = \frac{1}{2} \int (e^{e^{u}})' \ du$

$\displaystyle = \frac{1}{2} e^{e^{x^{2}}} + C$

**Krizalid** · January 26th 2011, 01:01 PM

perhaps this is one of the most boring integral bee, i remember old ones had much interesting problems.

15) you don't want to use product to sum formulae, we just put $x\mapsto \pi-x$ and that gives the same integral with a minus sign, since both are equal, then is obvious that the integral equals zero.

**simplependulum** · January 26th 2011, 11:30 PM

Having read the file of this test , i wonder whether it is possible to design a similar test representing this forum ( or we can just call it ' 2011 MHF Integration Bee Qualifying Test ' ) . In the test , I suggest proposing more difficult problems and to be more meaningful , I think all of the problems should be designed by our members ( it may be a difficult job ... or we can modify the old problems ) . All members are welcome to post their own problems , their problems will be put into the shortlist and finally 6 of them will be chosen for the test .

**Prove It** · January 26th 2011, 11:45 PM

1. $\displaystyle \int{\frac{x^6 - 1}{x^4 + x^3 - x - 1}\,dx} = \int{\frac{(x^3 - 1)(x^3 + 1)}{x^3(x + 1) - 1(x + 1)}\,dx}$

$\displaystyle = \int{\frac{(x^3 - 1)(x + 1)(x^2 - x + 1)}{(x + 1)(x^3 - 1) }\,dx}$

$\displaystyle = \int{x^2 - x + 1\,dx}$

$\displaystyle = \frac{x^3}{3} - \frac{x^2}{2} + x + C$ .

**simplependulum** · January 26th 2011, 11:50 PM

I have the answer for these two problems :

$\displaystyle \int \sqrt{\frac{x}{1-x^3} }~dx = -\frac{2}{3} \tan^{-1}\left( \sqrt{\frac{1}{x^3} - 1 }\right) + C$

Sub. $\displaystyle x = 1/t$

$\displaystyle -\int \sqrt{\frac{t^2}{t^3 - 1 } }~ \frac{dt}{t^2}$

$\displaystyle = - \int \frac{dt}{t\sqrt{t^3 - 1 }}$

Sub. $\displaystyle t^3 = u^2 + 1$ , $\displaystyle 3t^2 dt = 2u du$

$\displaystyle = -\frac{2}{3} \int \frac{du}{u^2 + 1 } = -\frac{2}{3} \tan^{-1}(u) + C$

$\displaystyle = -\frac{2}{3} \tan^{-1}\left( \sqrt{\frac{1}{x^3} - 1 } \right) + C$

This integral $\displaystyle I = \int_0^1 \frac{(x-1)^3}{(x+1)^4 }~dx$ is also interesting ,

sub. $\displaystyle x = \frac{1-t}{1+t}$ , $dx = -\frac{2}{(1+t)^2}~dt$

we have $\displaystyle I = - \int_0^1 \frac{t^3}{1 + \frac{1-t}{1+t} }~ \frac{2dt}{(1+t)^2}$

$\displaystyle = - \int_0^1 \frac{t^3}{1+t}~dt$

$\displaystyle = - \int_0^1 \frac{(1+t^3) - 1}{1+t}~dt$

$\displaystyle = \ln(2) - \frac{5}{6}$

**Prove It** · January 26th 2011, 11:56 PM

2. $\displaystyle \int{2\ln{x} + (\ln{x})^2\,dx} = \int{2\ln{x}\,dx} + \int{(\ln{x})^2\,dx}$

$\displaystyle = 2x\ln{x} - \int{\frac{2x}{x}\,dx} + \int{(\ln{x})^2\,dx}$

$\displaystyle = 2x\ln{x} - \int{2\,dx} + \int{(\ln{x})^2\,dx}$

$\displaystyle = 2x\ln{x} - 2x + \int{(\ln{x})^2\,dx}$

$\displaystyle = 2x\ln{x} - 2x + \int{(\ln{x})(\ln{x})\,dx}$

$\displaystyle = 2x\ln{x} - 2x + \ln{x}\left(x\ln{x} - x\right) - \int{\frac{x\ln{x} - x}{x}\,dx}$

$\displaystyle = 2x\ln{x} - 2x + x(\ln{x})^2 - x\ln{x} - \int{\ln{x} - 1\,dx}$

$\displaystyle = x\ln{x} - 2x + x(\ln{x})^2 - (x\ln{x} - x - x) + C$

$\displaystyle = x\ln{x} - 2x + x(\ln{x})^2 - x\ln{x} + 2x + C$

$\displaystyle = x(\ln{x})^2 + C$ .

**Prove It** · January 27th 2011, 12:00 AM

3. $\displaystyle \int{\frac{2x\,dx}{\sqrt{1 - x^4}}}$ .

Make the substitution $\displaystyle u = x^2$ so that $\displaystyle du = 2x\,dx$ and the integral becomes

$\displaystyle \int{\frac{du}{\sqrt{1 - u^2}}}$ .

Now make the substitution $\displaystyle u = \sin{\theta}$ so that $\displaystyle du = \cos{\theta}\,d\theta$ and the integral becomes

$\displaystyle \int{\frac{\cos{\theta}\,d\theta}{\sqrt{1 - \sin^2{\theta}}}}$

$\displaystyle = \int{\frac{\cos{\theta}\,d\theta}{\cos{\theta}}}$

$\displaystyle = \int{1\,d\theta}$

$\displaystyle = \theta + C$

$\displaystyle = \arcsin{u} + C$

$\displaystyle = \arcsin{(x^2)} +C$ .

**Prove It** · January 27th 2011, 12:04 AM

4. $\displaystyle \int{\frac{x^2 + 1}{x + 1}\,dx} = \int{\frac{x^2 + x - x + 1}{x + 1}\,dx}$

$\displaystyle = \int{\frac{x(x + 1)}{x + 1} + \frac{-x + 1}{x + 1}\,dx}$

$\displaystyle = \int{x + \frac{-x - 1 + 2}{x + 1} \,dx}$

$\displaystyle = \int{x + \frac{-(x + 1)}{x + 1} + \frac{2}{x + 1}\,dx}$

$\displaystyle = \int{x - 1 + \frac{2}{x + 1}\,dx}$

$\displaystyle = \frac{x^2}{2} - x + 2\ln{|x + 1|} + C$ .

**Prove It** · January 27th 2011, 12:13 AM

$\displaystyle \int{\frac{\sin^3{x} + \sin^2{x} - 2\sin{x} - 2}{\sin^2{x} + 2\sin{x} + 1}\,dx} = \frac{\sin^2{x}(\sin{x} + 1) - 2(\sin{x} + 1)}{(\sin{x} + 1)^2}$

$\displaystyle = \int{\frac{(\sin{x} + 1)(\sin^2{x} - 2)}{(\sin{x} + 1)^2}\,dx}$

$\displaystyle = \int{\frac{\sin^2{x} - 2}{\sin{x} + 1}}$

$\displaystyle = \int{\frac{\sin^2{x} + \sin{x} - \sin{x} - 2}{\sin{x} + 1}\,dx}$

$\displaystyle = \int{\frac{\sin{x}(\sin{x} + 1)}{\sin{x} + 1} +\frac{-\sin{x} - 2}{\sin{x} + 1}\,dx}$

$\displaystyle = \int{\sin{x} + \frac{-\sin{x} -1 -1}{\sin{x} + 1}\,dx}$

$\displaystyle = \int{\sin{x} - \frac{\sin{x} + 1}{\sin{x} + 1} - \frac{1}{\sin{x} + 1}\,dx}$

$\displaystyle = \int{\sin{x} - 1 - \frac{1}{\sin{x} + 1}\,dx}$

$\displaystyle = -\cos{x} - x - \int{\frac{1}{\sin{x} + 1}\,dx}$

$\displaystyle = -\cos{x} - x - \int{\frac{\sin{x} - 1}{(\sin{x} + 1)(\sin{x} - 1)}\,dx}$

$\displaystyle = -\cos{x} - x + \int{\frac{\sin{x} - 1}{\cos^2{x}}}$

$\displaystyle = -\cos{x} - x + \int{\frac{\sin{x}}{\cos^2{x}} - \sec^2{x}\,dx}$

$\displaystyle = -\cos{x} - x - \tan{x} - \int{u^{-2}\,du}$ after making the substitution $\displaystyle u = \cos{x}$

$\displaystyle = -\cos{x} - x - \tan{x} + u^{-1} + C$

$\displaystyle = \sec{x} -\cos{x} - x - \tan{x} + C$ .

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