Now its correct !! please share ur solution

# Thread to share Good Questions from Calculus which requires deep thinking

**pratyaksh_2019**#391

Put x^{2018} =t , then it converts to beta function B(a, b)

And then I think we can use the identity B(a, b) = \frac{ \Gamma (a) \Gamma (b) }{ \Gamma (a+b) }

**Tushar_2019**#392

@arush_kumar_singh only that options are given

@Rohan_Shinde1 , @pratyaksh_tyagi bro, solution is related to set theory and function

@Rohan_Shinde1 that's the beauty of question

**Rohan_Shinde1**#395

I can't think of a better way other than Beta function. Overkills the problem within 3 lines

**arush_2019**#396

@Tushar_Rathore

What is your solution bro?????

@Yash_Srivastava2

But can make questions cakewalk

**Rohan_Shinde1**#400

I still don't get it where is the set theory? I see function part overall but not set theory?

**arush_2019**#401

@Tushar_Rathore

I m not getting

How you prove that

\int _0^1 f(x) - f^{-1}(x) = 0

**Tushar_2019**#404

Try this one. This is an superb question

```
Find 2I-1
```

[.] This is only parenthesis not gif

**Rohan_Shinde1**#405

He applied the property that

\displaystyle \int_a^b f(x)dx +\int_{f(a)}^{f(b)} f^{-1}(x ) dx=bf(b)-af(a)