Finding an Integral from an Unknown Function

  • Thread starter Thread starter Khayyam89
  • Start date Start date
  • Tags Tags
    Function Integral
Khayyam89
Messages
7
Reaction score
0

Homework Statement


If xsin\pix = \intf(t) dt, where is a continuous function, find f(4).
b=x2, a=0

Homework Equations





The Attempt at a Solution


I assumed that the problem dealt with the Fundamental Theroem of Calculus so I began by saying that g(x)=xsin\pix but that is as far as it makes sense to me. Could anyone confirm to me that I am following the correct path?
 
Physics news on Phys.org
Yes, if
x sin(\pi x)= \int_0^{x^2} f(t)dt[/quote]<br /> you can apply the Fundamental theorem but if you do that, because the upper limit is &quot;x<sup>2</sup>&quot; rather than &quot;x&quot;, you need to use the chain rule also:<br /> F(x)= \int_0^x^2 f(t)dt= \int_0^u f(t)dt<br /> where u= x<sup>2</sup>. The derivative of the right hand side is F&#039;(u)(du/dx)= f(u)(2x)= f(x<sup>2</sup>)(2x). That is equal to the derivative of the left hand side. What is the derivative of x sin(\pi x)?<br /> f(
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top