For calculus lovers

Calculus Level 5

If f ( x ) = 1 + x 2 3 \displaystyle{f(x)~ =~ \sqrt [ 3 ]{ 1+x^2 } }

And g ( t ) = 0 t f ( x ) d x \displaystyle{g(t)~ =~ \int _{ 0 }^{ t }{ f(x)dx }}

Then find the value of ln ( tan ( π 6 ) ) ln ( tan ( π 3 ) ) g ( x ) d x \displaystyle{\int _{ \ln\left( \tan\left( \frac { \pi }{ 6 } \right) \right) }^{ \ln\left(\tan\left( \frac { \pi }{ 3 } \right) \right) }{ g(x)dx } } .


The answer is 0.00.

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1 solution

Aniket Verma
Apr 9, 2015

f ( x ) = 1 + x 2 3 \displaystyle{f(x)~ =~ \sqrt [ 3 ]{ 1+x^{ 2 }} }

f ( x ) = 1 + x 2 3 \displaystyle{ f(-x)~ =~ \sqrt [ 3 ]{ 1+x^{ 2 }} }

f ( x ) = f ( x ) ( t h e r e f o r e f ( x ) i s e v e n f u n c t i o n ) \displaystyle{ f(x)~ =~ f(-x)~ \cdots \cdots \cdots (therefore~ f(x)~ is~ even~ function)}

g ( t ) = 0 t f ( x ) d x ( t h e r e f o r e g ( t ) i s o d d f u n c t i o n ) \displaystyle{ g(t)~ =~ \int _{ 0 }^{ t }{ f(x)dx ~ \cdots \cdots \cdots \cdots (therefore~ g(t)~ is~ odd\quad function) }}

l e t l n ( t a n ( π 6 ) ) l n ( t a n ( π 3 ) ) g ( x ) d x = I \displaystyle{let~ \int _{ ln\left( tan\left( \frac { \pi }{ 6 } \right) \right) }^{ ln\left( tan\left( \frac { \pi }{ 3 } \right) \right) }{ g(x)dx~ =~ I }}

I = l n ( 3 ) l n ( 3 ) g ( x ) d x \displaystyle{\therefore ~ I~ =~ \int _{ -ln\left( \sqrt { 3 } \right) }^{ ln\left( \sqrt { 3 } \right) }{ g(x)dx }}

a a f ( x ) d x = 0 ( w h e r e f ( x ) i s a o d d f u n t i o n ) \displaystyle{\because ~ \int _{ -a }^{ a }{ f(x)dx~ =~ 0~ \cdots \cdots \cdots (where~ f(x)~ is~ a~ odd~ funtion) } }

I = 0 \therefore I~ =~ 0

yeah bro solved in the same way!!! nice problem BTW.

Shubhabrota Chakraborty - 6 years, 2 months ago

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