5 Challenging Integrals from the MIT Integration Bee
5 Challenging Integrals from the MIT Integration Bee

“The IITs became islands of excellence by not allowing the general debasement of the Indian system to lower their exacting standards. You couldn’t bribe your way to get into an IIT…Candidates are accepted only if they pass a grueling entrance exam.” The Wall Street Journal, on 16 April 2003 from WikiQuote.

These problems are based on questions from India’s JEE Main test. Students have about 1.5 minutes per question. While the test is multiple choice, I am presenting them without any options since the point is to understand how to solve them and rather than guessing the correct answer.

Test how quickly you can solve these integrals.

Problem 1

Problem 2

Problem 3

Watch the video for the solutions.

Solve In Just Seconds! (JEE Mains)

Or keep reading. “All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.
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Answer To Solve These Integrals In Seconds (IIT JEE Mains)

“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.

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(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Each of these problems can be solved with a clever technique summarized in this formula:

I credit the YouTube channel Vedantu JEE where I saw this formula. I have presented a proof later in this post. For now, let’s use the technique to solve the problems.

(And naturally the function f has to be well-behaved in the interval a to b. In the exam the functions are typically continuous in the integral, without any asymptotes, so you can use the formula without issue.)

Problem 1

Let f(x) = √x, so that f(2 + 4 – x) = f(6 – x) = √(6 – x). The above integral exactly fits the template of the formula, so we can the answer has to be (b – a)/2 = (4 – 2)/2 = 1.

And we’re done!

Problem 2

Let f(x) = √(sin x), so that f(0 + π2/ – x) = f(π/2 – x) = √(sin (π/2 – x)) = √(cos x). With this substitution, the above integral again fits the template of the formula, so we can the answer has to be (b – a)/2 = (π/2 – 0)/2 = π/4.

And like magic we have solved the problem!

Problem 3

We first need to do a substitution. Let u = x3, so that du = 3x2dx. We can substitute (1/3)du for x2dx. So then the integral takes the form:

Now we just need to notice:

log(3) + log(4) = log(3×4) = log(12)

Thus the above integral once fits the pattern and we can apply the formula. The integral then evaluates to (b – a)/2 = (log 4 – log 3)/2 = (1/2) log(4/3).

But we have to remember to multiply the integral by (1/3), so the answer is then (1/6) log(4/3).

That wasn’t so hard after all!

Proof

Suppose the integral has a value I. Now do a substitution with u = a + b – x so that du = –dx. After the change of variable, the integral is still equal to I, but notice its limits of integration are switched from b to a.

We can flip the limits of integration, which then cancels the -1 factor on du.

Then the variable u is a dummy variable, so we can change it to x.

Now we add both forms of the integral to get 2I is equal to the integral from a to b of the sum of those functions:

But the sum of the functions is (f(x) + f(a + b – x)/(f(x) + f(a + b – x) = 1. So we can integrate 1 from a to b:

This shows I = (b – a)/2, which concludes the proof.

Sources