This MIT Integral is Insultingly Fun But

[00:00:00]Okay, this integral looks almost insultingly easy. So we have 2 ^ of log of dx. So your brain says immediately that we have only dx and do integral say from 0 to 2 of x and dx that is equal to two except that the answer is wrong. So this is logarithm trap. So this whole problem depends on this small question that most of the people would never stop to ask. What is the base of the log? For the log, especially from math competition, we're using natural log. So for that log, we'll be using len of the x and do this integral, right? So let me just call this integral as the i.

[00:00:42]It's not the one. So your integral I is then going to be the same as integral from 0 to 2 of 2 ^ of ln of dx dx.

[00:00:55]Well, your first temptation was to come up with this a to the power of log the base of a with an argument of the x.

[00:01:02]This is equal to just the x, right?

[00:01:05]Well, take a look at this. If your integrant was 2 to the^ of log the base of two with an argument of the x, then this is equal to x, right? But then again, you have 2 to the^ of just log of dx. So, we're using natural log ln of dx.

[00:01:21]Okay? So what I'm going to do is I'll be just using this natural log, right? So we have integral is now 2 to the^ of ln of x.

[00:01:31]Okay. Then let's just use a to the power of b. a to the power of b is the same as then e to the power of b * ln of a.

[00:01:42]Okay, let's just use this right.

[00:01:45]So that 2 ^ of ln of x we can just rewrite this as e ^ of ln of x time ln of 2.

[00:02:00]Then we can just use how e to the power of say c * ln of x. Okay, this is the same as then x to the power of c. So using this your integrant which is 2 to the^ of ln of x it is just the same as x ^ of ln of 2 right so we can just use this x ^ of ln of two as your integrant okay and about this ln of two so ln of two the approximate value of the ln of two is around 0.69 693.

[00:02:40]So ln of 2 is around 0.693 and so on. This already gives useful information about the result because now then we'll be working on integrant of x ^ of 0.693.

[00:02:59]So for example when your x is between zero and one. Okay. If you're comparing x to the power of 0.693 and just the x, x to the power of 0.693 should be greater than the x, right?

[00:03:18]But then again, what if your x is between one and two? If your x is between one and two, then in this case, x has to be greater than x ^ of 0.693.

[00:03:30]So the result of this integral should be near two but not equal to two, right? So let's just double check.

[00:03:41]But then again this integral start with zero but ln of zero does not exist. So technically this has to be an improper integral. Right? So I'll be just rewriting this I integral I as now then the limit.

[00:03:58]Let me just use the absolute that goes to zero.

[00:04:03]Then we should have integral from epsilon to 2 of 2 ^ of ln of x dx.

[00:04:12]Okay. Then we already know 2 ^ of ln of x is the same as x ^ of ln of 2. We can just rewrite this as limit epsilon is going to zero of integral from epsilon to 2 of x ^ of ln dx.

[00:04:35]Okay, now everything looks really simple because integral say integral of x ^ of a dx. Okay, so this is pretty simple.

[00:04:44]This is same as x ^ of a + 1 over a + 1.

[00:04:50]So using this we can just evaluate this.

[00:04:52]Now in our case your a has to be ln of two.

[00:04:59]So that is why this integral of x ^ of ln2 dx. Okay, this is then going to be then the same as x ^ of 1 + ln2 that over 1 + ln2.

[00:05:18]Okay. So since we have the lower bound as the absolon and upper bound as the two. So we can just rewrite this I as limit epsilon is going to zero and then we have parenthesis uh plug it in 2 ^ of 1 + ln2 that over 1 + ln2 that minus epsilon the power of 1 + ln2 that over 1 + ln2.

[00:05:51]Okay, but then again ln of two is greater than zero, right?

[00:05:59]Ln of two is greater than zero. So that means 1 + ln of 2 should also be greater than zero.

[00:06:07]Then at the same time when your epsilon is going to zero then f of 1 + um ln2 okay should go to zero too. So that's why the second term will disappear. So that means the answer for this question is just 2 ^ of 1 + ln2 over 1 + ln2. This is the answer for the question.

[00:06:34]Okay, this was few suggested integral and we have a second method that is just a simple u substitution. You can substitute ln as the u and then everything else has to be simple. If you want to use the method and do this integral within two minutes, you can just go ahead and try. How amazing.