differentiating integrals

[colby_smith]

1. evaluate the integral g(x)=(1-t^2)^(1/3) from 1 to cos(x)

2.evaluate the integral y=sin(t^4) from 2x to 3x+1

on the first one i've made it this far,

u=1-t^2
-1/2du=tdt,
(u)^(1/3)-1/2du,

on the second,

u=t^4
(du)^(1/4) =tdt,
sin(u)du^(1/4)


show that the equation x^101+x^51+x-1=0 has exactly one real root

 

[tkhunny]

Just checking...Are you SURE you are being asked for definition integral. Maybe the derivative of the expression?

I might consider t = sin(u). I didn't check to see if it leads to anything.

On the last, do you know Descartes Rules of Signs? Perhaps you are being asked to use the Intermediate Value Theorem?

 

[raghvendrasingh]

let's suppose f(x)=x^101+x^51+x-1

now f(0)=-1

now f '(x)=101x^100+51x^50+1
since f '(x)>0 for all x therefore f(x) is an increasing function
and since f(0)=-1 the function f(x) will have only one real root.
Hence x^101+x^51+x-1=0 will have only one real root.

 

[Deleted member 4993]

Nice explanation Raghvendra....

 

[tkhunny]

let's suppose f(x)=x^101+x^51+x-1

now f(0)=-1

now f '(x)=101x^100+51x^50+1
since f '(x)>0 for all x therefore f(x) is an increasing function
and since f(0)=-1 the function f(x) will have only one real root.
Hence x^101+x^51+x-1=0 will have only one real root.

Not quite. You have demonstrated that there is at most one real root. You have not yet demonstrated that there actually is one. Apply this argument to f(x) = -1/x and see if you are convinced. Then complete the proof.

 

[raghvendrasingh]

Not quite. You have demonstrated that there is at most one real root. You have not yet demonstrated that there actually is one. Apply this argument to f(x) = -1/x and see if you are convinced. Then complete the proof.

since i have shown that the function f(x)=x^101+x^51+x-1 is an increasing function in its domain and also f(- infinity)=- infinity and f(+ infinity)=+ infinity and also f(x) is a continuous function in its domain since f(x) is a polynomial function .Hence as you move x from - infinity to + infinity the f(x) will also move from - infinity to + infinity and thus cutting the positive x axis only once since this is an increasing function in its domain and hence f(x) will have exactly one root .