2011 MIT Integration Bee Qualifying Test

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• Jan 27th 2011, 01:31 AM
Prove It
7. $\displaystyle \int{\sec^4{x}\tan^2{x}\,dx} = \int{\sec^2{x}\sec^2{x}\tan^2{x}\,dx}$.

Now using Integration by Parts with $\displaystyle u = \sec^2{x} \implies du = 2\tan{x}\sec^2{x}\,dx$ and $\displaystyle dv = \sec^2{x}\tan^2{x}\,dx \implies v = \frac{\tan^3{x}}{3}$ and the integral becomes

$\displaystyle \frac{\sec^2{x}\tan^3{x}}{3} - \int{\frac{2\tan^4{x}\sec^2{x}}{3}\,dx}$

$\displaystyle = \frac{\sec^2{x}\tan^3{x}}{3} - \frac{2}{3}\int{\tan^4{x}\sec^2{x}\,dx}$.

Now make the substitution $\displaystyle u = \tan{x} \implies du = \sec^2{x}\,dx$ and the integral becomes

$\displaystyle \frac{\sec^2{x}\tan^3{x}}{3} - \frac{2}{3}\int{u^4\,du}$

$\displaystyle = \frac{\sec^2{x}\tan^3{x}}{3} - \frac{2}{3}\left(\frac{u^5}{5}\right) + C$

$\displaystyle = \frac{\sec^2{x}\tan^3{x}}{3} - \frac{2\tan^5{x}}{15} + C$.
• Jan 27th 2011, 01:36 AM
Prove It
8. $\displaystyle \int{\sqrt{\csc{x} - \sin{x}}\,dx}= \int{\sqrt{\frac{1}{\sin{x}} - \sin{x}}\,dx}$

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

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

$\displaystyle = \int{\sqrt{\frac{\cos^2{x}}{\sin{x}}}\,dx}$

$\displaystyle = \int{\frac{\cos{x}}{\sqrt{\sin{x}}}\,dx}$

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

Now make the substitution $\displaystyle u = \sin{x} \implies du = \cos{x}\,dx$ and the integral becomes

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

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

$\displaystyle = 2\sqrt{\sin{x}} + C$.
• Jan 27th 2011, 01:58 AM
Prove It
9. $\displaystyle \int{\cos^6{x}\,dx} = \int{\left(\cos^2{x}\right)^3\,dx}$

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

$\displaystyle = \int{\frac{1}{8}\cos^3{2x} + \frac{3}{8}\cos^2{2x} + \frac{3}{8}\cos{2x} + \frac{1}{8}\,dx}$

$\displaystyle = \int{-\frac{1}{16} [-2\cos{2x}(1 - \sin^2{2x})] + \frac{3}{16} + \frac{3}{16}\cos{4x} + \frac{3}{8}\cos{2x} + \frac{1}{8}\,dx}$

$\displaystyle = \int{-2\cos{2x}\left(\frac{1}{16}\sin^2{2x} - \frac{1}{16}\right) + \frac{3}{16}\cos{4x} + \frac{3}{8}\cos{2x} + \frac{5}{16}\,dx}$

$\displaystyle = \int{\frac{1}{16}u^2 -\frac{1}{16}\,du} + \frac{3}{64}\sin{4x} + \frac{3}{16}\sin{2x} + \frac{5}{16}x + C$ after making the substitution $\displaystyle u = \sin{2x}$

$\displaystyle = \frac{1}{48}u^3 -\frac{1}{16}u + \frac{3}{64}\sin{4x} + \frac{3}{16}\sin{2x} + \frac{5}{16}x + C$

$\displaystyle = \frac{1}{48}\sin^3{2x} -\frac{1}{16}\sin{2x} + \frac{3}{64}\sin{4x} + \frac{3}{16}\sin{2x} + \frac{5}{16}x + C$

which I'm sure scrubs up much nicer than does here...
• Jan 27th 2011, 02:28 AM
Prove It
12. $\displaystyle \int{\frac{dx}{\cos{x}}} = \int{\frac{\cos{x}\,dx}{\cos^2{x}}}$

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

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

Now make the substitution $\displaystyle u = \sin{x} \implies du = \cos{x}\,dx$ and the integral becomes

$\displaystyle \int{\frac{du}{(1 - u)(1 + u)}}$.

Now using Partial Fractions:

$\displaystyle \frac{A}{1 - u} + \frac{B}{1 + u} = \frac{1}{(1 - u)(1 + u)}$

$\displaystyle \frac{A(1 + u) + B(1 - u)}{(1 - u)(1 + u)} = \frac{1}{(1 - u)(1 + u)}$

$\displaystyle A(1 + u) + B(1 - u) = 1$

$\displaystyle A + Au + B - Bu = 1$

$\displaystyle (A - B)u + A + B = 0u + 1$

$\displaystyle A-B = 0$ and $\displaystyle A + B = 1$

$\displaystyle A = B \implies A = \frac{1}{2}, B = \frac{1}{2}$.

So $\displaystyle \int{\frac{du}{(1 - u)(1 + u)}}= \int{\frac{1}{2(1 - u)} + \frac{1}{2(1 + u)}\,du}$

$\displaystyle = -\frac{1}{2}\ln{|1 - u|} + \frac{1}{2}\ln{|1 + u|} + C$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{1+u}{1 - u}\right|} + C$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{1 + \sin{x}}{1 - \sin{x}}\right|} + C$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{(1 + \sin{x})^2}{(1 - \sin{x})(1 + \sin{x})}\right|} + C$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{(1 + \sin{x})^2}{\cos^2{x}}\right|} + C$

$\displaystyle = \ln{\left|\left(\frac{1 + \sin{x}}{\cos{x}}\right)^2\right|^{\frac{1}{2}}} + C$

$\displaystyle = \ln{\left|\frac{1 + \sin{x}}{\cos{x}}\right|} + C$

$\displaystyle = \ln{|\sec{x} + \tan{x}|} + C$.
• Jan 27th 2011, 02:35 AM
Prove It
17. $\displaystyle \int{\frac{dx}{2 + e^x}} = \int{\frac{e^x\,dx}{2e^x + e^{2x}}}$.

Now let $\displaystyle u = e^x \implies du = e^x\,dx$ and the integral becomes

$\displaystyle \int{\frac{du}{2u + u^2}}$

$\displaystyle = \int{\frac{du}{u(2 + u)}}$.

Now using Partial Fractions:

$\displaystyle \frac{A}{u} + \frac{B}{2 + u} = \frac{1}{u(2 + u)}$

$\displaystyle \frac{A(2 +u) + Bu}{u(2 + u)} = \frac{1}{u(2 + u)}$

$\displaystyle A(2 + u) + Bu = 1$

$\displaystyle 2A + Au + Bu = 1$

$\displaystyle (A + B)u + 2A = 0u + 1$

$\displaystyle A + B = 0$ and $\displaystyle 2A = 1$

$\displaystyle A = \frac{1}{2}$ and $\displaystyle B = -\frac{1}{2}$.

So $\displaystyle \int{\frac{du}{u(2+u)}} = \int{\frac{1}{2u} - \frac{1}{2(2 + u)}\,du}$

$\displaystyle = \frac{1}{2}\ln{|u|} - \frac{1}{2}\ln{|2+u|} + C$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{u}{2 + u}\right|} + C$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{e^x}{2 + e^x}\right|} + C$
• Jan 27th 2011, 02:41 AM
Prove It
19. $\displaystyle \int{\frac{4x\,dx}{1 - x^4}} = 2\int{\frac{2x\,dx}{1 - (x^2)^2}}$.

Let $\displaystyle u = x^2 \implies du = 2x\,dx$ and the integral becomes

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

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

$\displaystyle = 2\int{\frac{1}{2(1 + u)} + \frac{1}{2(1 - u)}\,du}$ after using Partial Fractions

$\displaystyle = 2\left(\frac{1}{2}\ln{|1 + u|} - \frac{1}{2}\ln{|1-u|}\right) + C$

$\displaystyle = \ln{|1 + u|} - \ln{|1 - u|} + C$

$\displaystyle = \ln{\left|\frac{1+u}{1 - u}\right|} + C$

$\displaystyle = \ln{\left|\frac{1 + x^2}{1 - x^2}\right|} + C$.
• Jan 27th 2011, 02:43 AM
Ted
Prove It:
for #17, If you multiply and devide by $e^{-x}$ it will be solved in 1 second.
• Jan 27th 2011, 02:55 AM
Prove It
21. $\displaystyle \int_0^6{\sqrt{6x - x^2}\,dx} = \int_0^6{\sqrt{-(x^2 - 6x)}\,dx}$

$\displaystyle = \int_0^6{\sqrt{-\left[x^2 - 6x + (-3)^2 - (-3)^2\right]}\,dx}$

$\displaystyle = \int_0^6{\sqrt{-\left[(x - 3)^2 - 9\right]}\,dx}$

$\displaystyle = \int_0^6{\sqrt{9 - (x - 3)^2}\,dx}$.

Note that $\displaystyle y = \sqrt{9 - (x - 3)^2}$ is a semicircle centred at $\displaystyle (3,0)$ of radius $\displaystyle 3$.

So this integral is the area of that semicircle.

$\displaystyle \int_0^6{\sqrt{9-(x-3)^2}\,dx} = \frac{1}{2}\cdot \pi \cdot 3^2$

$\displaystyle = \frac{9\pi}{4}$.
• Jan 27th 2011, 03:12 AM
Prove It
$\displaystyle \int_0^1{\frac{x^3 - 3x^2 + 3x - 1}{x^4 + 4x^3 + 6x^2 + 4x + 1}\,dx} = \int_0^1{\frac{(x - 1)^3}{(x + 1)^4}}$.

Now let $\displaystyle u = x+1 \implies x - 1 = u - 2$ and $\displaystyle du = dx$.

When $\displaystyle x = 0, u = 1$ and when $\displaystyle x = 1, u = 2$.

The integral becomes

$\displaystyle \int_1^2{\frac{(u-2)^3}{u^4}\,du}$

$\displaystyle = \int_1^2{\frac{u^3 - 6u^2 + 12u - 8}{u^4}\,du}$

$\displaystyle = \int_1^2{u^{-1} - 6u^{-2} + 12u^{-3} - 8u^{-4}\,du}$

$\displaystyle = \left[\ln{|u|} + 6u^{-1} - 6u^{-2} + \frac{8u^{-3}}{3}\right]_1^2$

$\displaystyle = \left(\ln{|2|} + 3 - \frac{3}{2} + \frac{1}{3}\right) - \left(\ln{|1|} + 6 - 6 + \frac{8}{3}\right)$

$\displaystyle = \ln{(2)} - \frac{5}{6}$.
• Jan 27th 2011, 03:20 AM
Prove It
25. $\displaystyle \int{\sqrt{\frac{1-x}{1+x}}\,dx} = \int{\sqrt{\frac{(1 - x)(1-x)}{(1 + x)(1 - x)}}\,dx}$

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

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

Now let $\displaystyle x = \sin{\theta} \implies dx = \cos{\theta}\,d\theta$ and the integral becomes

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

$\displaystyle = \int{\frac{(1 - \sin{\theta})\cos{\theta}}{\cos{\theta}}\,d\theta}$

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

$\displaystyle = \theta + \cos{\theta} + C$

$\displaystyle = \theta+ \sqrt{1 - \sin^2{\theta}} + C$

$\displaystyle = \arcsin{x}+ \sqrt{1 - x^2} + C$.
• Jan 27th 2011, 03:26 AM
Prove It
23. $\displaystyle \int{x\,e^{e^{x^2} + x^2}\,dx} = \frac{1}{2}\int{e^{e^{x^2} + x^2}\cdot 2x\,dx}$.

Let $\displaystyle u = x^2 \implies du = 2x\,dx$ and the integral becomes

$\displaystyle \frac{1}{2}\int{e^{e^u + u}\,du}$

$\displaystyle = \frac{1}{2}\int{e^{e^u}\cdot e^u\,du}$.

Now let $\displaystyle v = e^u \implies dv = e^u\,du$ and the integral becomes

$\displaystyle \frac{1}{2}\int{e^v\,dv}$

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

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

$\displaystyle = \frac{1}{2}e^{e^{x^2}} + C$.
• Jan 27th 2011, 03:46 AM
Prove It
10. $\displaystyle \int{\frac{dx}{1 + 2x^2 + x^4}} = \int{\frac{dx}{(1 + x^2)^2}}$

Let $\displaystyle x = \tan{\theta} \implies dx = \sec^2{\theta}\,d\theta$ and the integral becomes

$\displaystyle \int{\frac{\sec^2{\theta}\,d\theta}{(1 + \tan^2{\theta})^2}}$

$\displaystyle = \int{\frac{\sec^2{\theta}\,d\theta}{\sec^4{\theta} }}$

$\displaystyle = \int{\frac{d\theta}{\sec^2{\theta}}}$

$\displaystyle = \int{\cos^2{\theta}\,d\theta}$

$\displaystyle = \int{\frac{1}{2} + \frac{1}{2}\cos{2\theta}\,d\theta}$

$\displaystyle =\frac{1}{2}\theta + \frac{1}{4}\sin{2\theta} + C$

$\displaystyle =\frac{1}{2}\theta + \frac{1}{4}\left(\frac{2\tan{\theta}}{1 + \tan^2{\theta}}\right) + C$

$\displaystyle = \frac{1}{2}\theta + \frac{\tan{\theta}}{2(1 + \tan^2{\theta})} + C$

$\displaystyle = \frac{1}{2}\arctan{x} + \frac{x}{2(1 + x^2)} + C$.
• Jan 27th 2011, 03:53 AM
Prove It
Just out of interest, having never been in an Integral Bee before, how many do you have to get right in the time limit to qualify?
• Jan 27th 2011, 05:24 AM
TheCoffeeMachine
Quote:

Originally Posted by Prove It
Just out of interest, having never been in an Integral Bee before, how many do you have to get right in the time limit to qualify?

Don't know, I'm wondering that too. It might depend on the boundaries of that particular year. I think with sufficient training and wise choice of questions, most people can get more than ten right within the time limit. But you have to keep in mind that they are not doing them in the regular way -- with the explicit substitutions and all. For example, it's much more efficient to do # 10 as follows:

\begin{aligned} & \int{\sqrt{\frac{1-x}{1+x}}\,dx} = \int{\frac{1 - x}{\sqrt{1 - x^2}}\,dx} \\& = \int \frac{1}{\sqrt{1-x^2}}\;{dx}-\int\frac{(x^2)'}{2\sqrt{1-x^2}}\;{dx} \\& \sin^{-1}{x}+\sqrt{1-x^2}+k. \end{aligned}
• Jan 27th 2011, 05:47 AM
TheCoffeeMachine
By the way, here is a head-to-head video of the final round (not sure which year).
It took the winner about 30 seconds to do $\int \frac{x^4}{1-x^2}\;{dx}$. What do you think?
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