Indefinite integral w/ continous functions: f(x)={ 2x-1,x<=1} { x^2,x>1}

[wolly]

I have f(x)={ 2x-1,x<=1}
{ x^2,x>1}
Now,after this the problem requires me to find the antiderivative of f(x) which is F(x)(the primitive function).
F(x)={x^2-x,x<=1}
{x^3/3,x>1}
If the integral of f(x) is F(x) then the derivative of F(x) is f(x)(the original function).I stated this because I didn't want to create a confusion.
The problem also requires me to check if the values of the primitive function F(x) is F(4)=-8.I tried to plug the values into the function F but I couldn't get the -8 value.How do I solve this problem?Also the problem wants me to find F(-1) and the answer to this is -27.How can F(-1)=-27?

 

[Dr.Peterson]

I have f(x)={ 2x-1,x<=1}
{ x^2,x>1}
Now,after this the problem requires me to find the antiderivative of f(x) which is F(x)(the primitive function).
F(x)={x^2-x,x<=1}
{x^3/3,x>1}
If the integral of f(x) is F(x) then the derivative of F(x) is f(x)(the original function).I stated this because I didn't want to create a confusion.
The problem also requires me to check if the values of the primitive function F(x) is F(4)=-8.I tried to plug the values into the function F but I couldn't get the -8 value.How do I solve this problem?Also the problem wants me to find F(-1) and the answer to this is -27.How can F(-1)=-27?

I think you've misunderstood the role of F(4)=-8 in the problem. It is not just a check.

Any function has many antiderivatives, not just one. They vary in the arbitrary constant, often called C, that must be added. What you have found is AN antiderivative for each case, not THE antidevative.

They are probably asking you to find the antiderivative for which F(4)=-8. So you need to add a C to each case of your F, use the given condition to find the value of one of the C's, and then find the value of the other C so that the function is continuous. (Your F is not.)

Then, after all that, you can find F(-1).

If you are not sure that I have interpreted the problem correctly, please do as we ask and tell us the exact wording of the entire problem as given to you. That can make a big difference.

 

[wolly]

And how do I solve it?

 

[Dr.Peterson]

And how do I solve it?

I think I told you what to do. Please try, and show me your work. (Presumably you have been taught what to do, as well.)

The first step, as I said, is to write your antiderivatives with arbitrary constants:

F(x) = {x^2 - x + C₁, x<=1}
{x^3/3 + C₂, x>1}
​

Now use the given condition, that F(4) = -8, to determine one of the constants. Which will it be? What is its value?

Then go as far as you can beyond that, and show your work to get the next hint.

 

[wolly]

I think I told you what to do. Please try, and show me your work. (Presumably you have been taught what to do, as well.)

The first step, as I said, is to write your antiderivatives with arbitrary constants:

F(x) = {x^2 - x + C₁, x<=1}
{x^3/3 + C₂, x>1}
​

Now use the given condition, that F(4) = -8, to determine one of the constants. Which will it be? What is its value?

Then go as far as you can beyond that, and show your work to get the next hint.

https://postimg.cc/rdYP8gPj/b0215f94
Did I solved it correctly?

 

[Dr.Peterson]

Did I solved it correctly?

Please don't post links to that site, which is forbidden.

Here is your image:



Your answer is

F(x) = { x^2 - x - 8, x <= 1
.........{ (x^3 - 25)/3, x > 1
​

Let's check. F(4) = (4^3 - 25)/3 = (64 - 25)/3 = 13, not -8, so this is wrong.

You used the wrong formula to find your first constant.

How about continuity?

lim{x->1-} F(x) = 1^2 - 1 - 8 = -8
lim{x->1+} F(x) = (1^3 - 25)/3 = -8
​

so this is okay. But you'll have to redo this when you correct F(4).

 

[wolly]

Your answer is

F(x) = { x^2 - x - 8, x <= 1
.........{ (x^3 - 25)/3, x > 1
​

​

​

Ok so now I know what F(4)=-8 means.If I plug in the value 4 in the x^3/3+c2 I get 64/3+c2=-8 and c2=-88/3.I found the constant c1=-8.How does that help me to find F(-1)?

 

[HallsofIvy]

-1< 1 isn't it? And Dr. Peterson told you that, if x< 1, \(\displaystyle F)(x)= x^2- x- 8\). So F(-1)= (-1)^2- (-1)- 8.

 

[wolly]

-1< 1 isn't it? And Dr. Peterson told you that, if x< 1, \(\displaystyle F)(x)= x^2- x- 8\). So F(-1)= (-1)^2- (-1)- 8.

In my textbook it says the answer is F(-1)=-27.

 

[Dr.Peterson]

[/INDENT]Ok so now I know what F(4)=-8 means.If I plug in the value 4 in the x^3/3+c2 I get 64/3+c2=-8 and c2=-88/3.I found the constant c1=-8.How does that help me to find F(-1)?

You'll have to put this C₂ into the second case, and then find C₁ to make the function continuous. (It is not -8 this time.) THEN you can evaluate F(-1). (When I do this, I do get -27.)

 

[wolly]

You'll have to put this C₂ into the second case, and then find C₁ to make the function continuous. (It is not -8 this time.) THEN you can evaluate F(-1). (When I do this, I do get -27.)

And how do I do that?My textbook says nothing about these problems.

 

[topsquark]

You'll have to put this C₂ into the second case, and then find C₁ to make the function continuous. (It is not -8 this time.) THEN you can evaluate F(-1). (When I do this, I do get -27.)

I have looked at this problem on MMF as well and on neither this site nor that does it stipulate that F(x) should be continuous, though that you achieved the correct result says that it's likely.

-Dan

 

[pka]

I have \(\displaystyle f(x)=\begin{cases}2x-1&, x\le1 \\ x^2 &, x>1\end{cases}\)
Now,after this the problem requires me to find the antiderivative of f(x) which is F(x)(the primitive function).
\(\displaystyle F(x)=\begin{cases}x^2-x+C_1 &, x\le1 \\ \dfrac{x^3}{3}+C_2 &, x>1\end{cases}\)
If the integral of f(x) is F(x) then the derivative of F(x) is f(x)(the original function).I stated this because I didn't want to create a confusion.
The problem also requires me to check if the values of the primitive function F(x) is F(4)=-8.I tried to plug the values into the function F but I couldn't get the -8 value.How do I solve this problem?Also the problem wants me to find F(-1) and the answer to this is -27.How can F(-1)=-27?

Now find \(\displaystyle C_1~\&~C_2\)
\(\displaystyle -8=F(4)=\dfrac{4^3}{3}+C_2\) so that \(\displaystyle -8=\dfrac{64}{3}+C_2\) or \(\displaystyle C_2=-\dfrac{88}{3}\)
\(\displaystyle \mathop {\lim }\limits_{x \to {1^ - }} \left( {{x^2} - x + {C_1}} \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {\dfrac{{{x^3}}}{3} - \dfrac{{88}}{3}} \right)\) so \(\displaystyle C_1=-29\)
FINISH??

 

[Dr.Peterson]

I have looked at this problem on MMF as well and on neither this site nor that does it stipulate that F(x) should be continuous, though that you achieved the correct result says that it's likely.

-Dan

Well, the thread title is " Indefinite integral with continous fun[c]tions problem"; and in order to have the stated (continuous) derivative f(x), F(x) has to be continuous ...

But it would have been nice if we were given the actual statement of the problem.