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December 14th, 2018, 04:20 AM  #1 
Newbie Joined: Aug 2016 From: Romania Posts: 28 Thanks: 1  Indefinite integral with continous functions problem
I have f(x)={ 2x1,x<=1} { x^2,x>1} I tried to use the indefinite integral formula to find the primitive function F(x) and this is what I got F(x)={x^2x,x<=1} {x^3/3,x>1} The problem wants to check if F(4)=8 and to find F(1).How do I find F(1)?Do I use limits? 
December 14th, 2018, 04:24 AM  #2 
Senior Member Joined: Oct 2009 Posts: 850 Thanks: 326 
Hint: a function has multiple primitives. How do you know which one to take?

December 14th, 2018, 06:45 AM  #3  
Newbie Joined: Aug 2016 From: Romania Posts: 28 Thanks: 1  Quote:
How do I find F(1) from the values of the constants?  
December 14th, 2018, 09:01 AM  #4 
Newbie Joined: Aug 2016 From: Romania Posts: 28 Thanks: 1 
I forgot to mention that F(1)=27.How can I prove this?

December 14th, 2018, 09:51 AM  #5  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 928 Math Focus: Wibbly wobbly timeywimey stuff. 
You aren't proving anything. Your problem statement is likely: Quote:
$\displaystyle F(x) = \int f(x) ~ dx = \begin{cases} x^2  x + C_1 & x \leq 1 \\ \dfrac{x^3}{3} + C_2 & 1 < x \end{cases}$ F(4) gives a value for $\displaystyle C_2$ and F(1) gives you a value for $\displaystyle C_1$. Dan  
December 14th, 2018, 09:58 AM  #6 
Newbie Joined: Aug 2016 From: Romania Posts: 28 Thanks: 1 
@topsquark I have to find F(1) given that F(4)=8.How do I find it?

December 14th, 2018, 10:03 AM  #7  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 928 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Now, if there is an extra feature such as that F(x) is continuous (which seemed to be assumed by the workers on FMH) then you can say that $\displaystyle F(1) = 1^2  1 + C_1 = \dfrac{1^3}{3}  \dfrac{88}{3}$ then you can solve for $\displaystyle C_1$ and then just plug in F(1). But as the problem statement stands the problem can't be done. That's why I proposed the problem statement above as I did. Dan  
December 14th, 2018, 11:18 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,927 Thanks: 2205 
You need $C_2 = 88/3$ so that $4^3\!/3  88/3 = 24/3 = 8$ (as given). As $F(x)$ must be continuous at $x = 1$, $1^2  1 + C_1 = 1^3/3  88/3 = 29$, and so $C_1 = 29$. $(1)^2 (1)  29 = 1 + 1  29 = 27$. 

Tags 
continous, functions, funtions, indefinite, integral, problem 
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