A Tricky Integral for some

Calculus Level 5

lim n 0 1 ( x n + ( 1 x ) n ) 1 n d x = ? \large \lim_{n \to \infty} \int^{1}_{0} \left(x^{n}+(1-x)^{n} \right)^{\frac{1}{n}}dx = \ ?

This problem is a part of my set Some JEE problems

3 4 \frac{3}{4} 1 3 2 \frac{3}{2} 0 3 8 \frac{3}{8}

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1 solution

Tanishq Varshney
Apr 27, 2015

Ishan Dasgupta Samarendra here is the solution

using the property f ( x ) = f ( 2 a x ) f(x)=f(2a-x) the limit becomes half

2 0 1 2 ( x n + ( 1 x ) n ) 1 n d x 2 \displaystyle \int^{\frac{1}{2}}_{0} (x^{n}+(1-x)^{n})^{\frac{1}{n}} dx ..... ( 1 ) (1)

plz see carefully.

for x ( 0 , 1 2 ) x\in (0,\frac{1}{2})

x x has smaller value than ( 1 x ) (1-x) .

taking common ( 1 x ) n (1-x)^{n} in ( 1 ) (1)

also lim n ( x 1 x ) n 0 \displaystyle \lim_{n\to \infty} (\frac{x}{1-x})^{n} \rightarrow 0

the integral reduces to

2 0 1 2 ( 1 x ) × ( 1 n ) 1 n . d x 2 \displaystyle \int^{\frac{1}{2}}_{0} (1-x) \times (1^{n})^{\frac{1}{n}}.dx

2 0 1 2 ( 1 x ) . d x 2 \displaystyle \int^{\frac{1}{2}}_{0} (1-x).dx

3 4 \frac{3}{4}

hope it helps

You cannot take the limit inside so easily. You'll have to use Dominated Convergence Theorem showing that it is less than/equal to some integrable function for all n. And then you can go as you did.

Kartik Sharma - 5 years, 9 months ago

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You know, as time passes, I just keep marveling at your grasp of Calculus.. When I was your age, all I would do was play Badminton across the country and dream of representing the nation on the International stage!

User 123 - 5 years, 9 months ago

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All I can say is you had a great teen life. At least you had a dream(and that too so big). "Sports is what my body miss and failure is what my mind piss" :( :/. You have had a great life before and a greater life ahead. Cheer up man!

Kartik Sharma - 5 years, 9 months ago

Nice question!!...took time to solve it...but it is not a JEE type question...it uses undergraduate level mathematics

Ravi Dwivedi - 4 years, 11 months ago

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