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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)={ 2x-1,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^2-x,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? Thanks from topsquark
December 14th, 2018, 06:45 AM   #3
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Quote:
 Originally Posted by Micrm@ss Hint: a function has multiple primitives. How do you know which one to take?
I calculated F(4)=-8 and I got from the 2 constants(which I forgot to write)and now I have c1=-8 from x<=1 and from x>1 4^3/3+c2=-8.Ok now c2=-88/3.
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
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You aren't proving anything. Your problem statement is likely:
Quote:
 Given $\displaystyle f(x) = \begin{cases} 2x - 1 & x \leq 1 \\ x^2 & 1 < x \end{cases}$ Find $\displaystyle F(x) = \int f(x)~dx$ where F(4) = -8 and F(-1) = -27.
The F(4) and F(1) conditions give you the values of your constants. eg.

$\displaystyle F(x) = \int f(x) ~ dx = \begin{cases} x^2 - x + C_1 & x \leq 1 \\ \dfrac{x^3}{3} + C_2 & 1 < x \end{cases}$

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
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Quote:
 Originally Posted by alex77 @topsquark I have to find F(-1) given that F(4)=-8.How do I find it?
If you have posted the problem as stated then there is no way. I note that you have posted the same thing on Free Math Help. If there is any more information in the image you posted I couldn't read it as I couldn't blow up the image.

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 $C_1 = -29$. $(-1)^2 -(-1) - 29 = 1 + 1 - 29 = -27$. Thanks from topsquark

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