I often used to remember a mathematical problem [from IIT entrance test papers], which took me a lot of days to solve in my school level.
The problem is to find the value of the following definite integral, $$ I = 5050 \frac{\int_0^1 (1-x^{50})^{100} \,dx }{\int_0^1(1-x^{50})^{101}\,dx}$$
Giving some time before going to the solution, gives you enough insight into the problem.
Of course I will give you the answer but not solution because the answer is what surprised more than anything else. The value of above integral is - 5051.
I took this problem just because of this answer because,
here 5050 * {Integral} = 5051
Simply the integral is so amazing that it just adds the amount by one number when multiplied!!!
For the solution,
Most of the time, we tend to solve the integral using all our pre-loaded formulas in our mind. I myself tend to solve this using many possible methods including series of expansion, integral by parts, even tried find in a numerical way, etc.
We always used to stick only with the given terms in the problem, but that won't help here in anyway
So, you need to change the power of the integral but need to keep the integral unchanged. Perfect choice is multiply and divide by same value.
Again it won't help with the integral since it would change power both in the numerator and the denominator.
Note:You need to multiply only in the numerator.
The choice would be the number - "1"
Yet since we also need $x^{50}$ here the best choice would be $$ 1- x^{50}+x^{50}$$
Thus we never changed any value but multiplied a term only in the numerator to change the power.
The rest of the problem is just simple mathematics as follows,
$$ I = 5050 \left(\frac{\int_0^1 (1-x^{50})^{100} \times (1-x^{50}+x^{50}) \,dx }{ \int_0^1(1-x^{50})^{101}\,dx }\right)$$
which gives,
$$ I = 5050 \left( \frac{ \int_0^1 (1-x^{50})^{101}\,dx }{ \int_0^1 (1-x^{50})^{101}\,dx } + \frac {\int_0^1 (1-x^{50})^{100}x^{50} \,dx }{ \int_0^1 (1-x^{50})^{101} \,dx}\right) \\~\\ I = 5050 \left( 1 + \frac{\int_0^1 x^{50}\,dx}{\int_0^1 (1-x^{50})^{101}\,dx}\right)$$
The integral on applying the limits simplifies into...
$$ I = 5050 \left( 1 + \frac{\int_0^1(1-x^{50}) x^{50}\,dx}{\int_0^1(1-x^{50})^{101}\,dx}\right)$$
We will do the separately the integral in the denominator using "integration by parts method"..
$$ I_0 =\int_0^1(1-x^{50})^{101}\,dx = [x(1-x^{50})^{101}]|_0^1 - \int_0^1 (x) (101) (-50x^{49}) (1-x^{50})^{100} \,dx \\~\\ I_0 = 0 + (5050) \int_0^1(1-x^{50})^{100}x^{50} /,dx $$
Thus we got.. $$ I_0 = (5050) \int_0^1 (1-x^{50})^{100} x^{50} \,dx $$The problem is to find the value of the following definite integral, $$ I = 5050 \frac{\int_0^1 (1-x^{50})^{100} \,dx }{\int_0^1(1-x^{50})^{101}\,dx}$$
Giving some time before going to the solution, gives you enough insight into the problem.
Of course I will give you the answer but not solution because the answer is what surprised more than anything else. The value of above integral is - 5051.
I took this problem just because of this answer because,
here 5050 * {Integral} = 5051
Simply the integral is so amazing that it just adds the amount by one number when multiplied!!!
For the solution,
Most of the time, we tend to solve the integral using all our pre-loaded formulas in our mind. I myself tend to solve this using many possible methods including series of expansion, integral by parts, even tried find in a numerical way, etc.
We always used to stick only with the given terms in the problem, but that won't help here in anyway
Problem is all about a simple trick, which you can understand by just looking at the numbers i.e. 100 and 101
To solve it, we should simplify the integral but it is possible only when they both have the same powers. So, you need to change the power of the integral but need to keep the integral unchanged. Perfect choice is multiply and divide by same value.
Again it won't help with the integral since it would change power both in the numerator and the denominator.
Note:You need to multiply only in the numerator.
The choice would be the number - "1"
Yet since we also need $x^{50}$ here the best choice would be $$ 1- x^{50}+x^{50}$$
Thus we never changed any value but multiplied a term only in the numerator to change the power.
The rest of the problem is just simple mathematics as follows,
$$ I = 5050 \left(\frac{\int_0^1 (1-x^{50})^{100} \times (1-x^{50}+x^{50}) \,dx }{ \int_0^1(1-x^{50})^{101}\,dx }\right)$$
which gives,
$$ I = 5050 \left( \frac{ \int_0^1 (1-x^{50})^{101}\,dx }{ \int_0^1 (1-x^{50})^{101}\,dx } + \frac {\int_0^1 (1-x^{50})^{100}x^{50} \,dx }{ \int_0^1 (1-x^{50})^{101} \,dx}\right) \\~\\ I = 5050 \left( 1 + \frac{\int_0^1 x^{50}\,dx}{\int_0^1 (1-x^{50})^{101}\,dx}\right)$$
The integral on applying the limits simplifies into...
$$ I = 5050 \left( 1 + \frac{\int_0^1(1-x^{50}) x^{50}\,dx}{\int_0^1(1-x^{50})^{101}\,dx}\right)$$
We will do the separately the integral in the denominator using "integration by parts method"..
$$ I_0 =\int_0^1(1-x^{50})^{101}\,dx = [x(1-x^{50})^{101}]|_0^1 - \int_0^1 (x) (101) (-50x^{49}) (1-x^{50})^{100} \,dx \\~\\ I_0 = 0 + (5050) \int_0^1(1-x^{50})^{100}x^{50} /,dx $$
substituting this in "I" .. we arrive at..
$$ I = 5050 \left( 1 + \frac{1}{5050} \frac {\int_0^1 (1-x^{50})^{100} x^{50} \,dx}{\int_0^1 (1-x^{50})^{100} x^{50} \,dx}\right)$$
$$ \rightarrow I = 5050 \left( 1 + \frac{1}{5050} \right) $$
Thus, we finally arrived at our amazing solution
I = 5050 + 1 = 5051
Now, we can go and look at the beauty of the solution!!!
