Winter Integers

Calculus Level 4

Let f : R R f : \mathbb R \rightarrow \mathbb R be a continuous function with

0 1 f ( x ) f ( x ) d x = 0 0 1 ( f ( x ) ) 2 f ( x ) d x = 18 \large\ { \displaystyle \int _{ 0 }^{ 1 }{ f\left( x \right) f'\left( x \right) dx } = 0 \\ \displaystyle \int _{ 0 }^{ 1 }{ { \left( f\left( x \right) \right) }^{ 2 } } f'\left( x \right) dx = 18 } .

Find 0 1 ( f ( x ) ) 4 f ( x ) d x \large\ \displaystyle \int _{ 0 }^{ 1 }{ { \left( f\left( x \right) \right) }^{ 4 }f'\left( x \right) dx } .


The answer is 97.2.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

X X
May 26, 2018

( f ( x ) ) 2 0 1 = 2 0 1 f ( x ) f ( x ) d x = 0 \left.(f(x))^2\right|_0^1=2\int_{ 0 }^{ 1 }{ f\left( x \right) f'\left( x \right) dx } = 0 ( f ( x ) ) 3 0 1 = 3 0 1 ( f ( x ) ) 2 f ( x ) d x = 54 \left.(f(x))^3\right|_0^1=3\int_{ 0 }^{ 1 }{ (f\left( x \right))^2 f'\left( x \right) dx } = 54 f 2 ( 1 ) = f 2 ( 0 ) , f 3 ( 1 ) f 3 ( 0 ) = 54 f ( 1 ) = f ( 0 ) , 2 f 3 ( 1 ) = 54 , f ( 1 ) = 3 f^2(1)=f^2(0),f^3(1)-f^3(0)=54\rightarrow f(1)=-f(0),2f^3(1)=54,f(1)=3 ( f ( x ) ) 5 0 1 = 5 0 1 ( f ( x ) ) 4 f ( x ) d x = 2 × 3 5 = 486 \left.(f(x))^5\right|_0^1=5\int_{ 0 }^{ 1 }{ (f\left( x \right))^4 f'\left( x \right) dx } = 2\times3^5=486 0 1 ( f ( x ) ) 4 f ( x ) d x = 486 5 \int _{ 0 }^{ 1 }{ { \left( f\left( x \right) \right) }^{ 4 }f'\left( x \right) dx } =\frac{486}5

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...