Please explain this calculus solution

[barryj]

Homework Statement:

I have this problem and solution. I do not understand the solution, in particular how the integral fits into the solution. Is this a form of the 2nd theorem of calculus perhaps? I am confused.
if f'(x) = cos (x^3) and f(1) = 2 find f(0)

Relevant Equations:

I have attached the problem statement and he solution.

If f'(x) were a simpler function like f'(x) = cos(x) I would say

f(x) = sin(x) + C and then evaluate C by knowing that 2 = sin(1) + C and then C would equal 2-sin(1)
the f(x) = sin(x) + 2 - sin(1),
f(0) = sin(0) + 2 - sin(1) = 0 + 2 -.841 = 1.58
However the more complicated problem has f'(x) - cos(x^3) and it is solved using an integral expression that is evaluated with a calculator.
I do not understand this integral method. Please help or point me to a link where I can find out what is going on.

 

[FactChecker]

They don't really figure out the integral, so they leave the integral there. This is more about how to use the fact that f(1)=2. They start their integral at x=1 so that the integral value is 0 there and the total of the integral and constant is 2. Then, to get the integral value at 0, they integrate the (unsolved) integral down to x=0. That is their final answer. It is correct and accurate, but the numerical value is unknown because they never really solve the integral.

EDIT: Sorry, they do get a numerical value in the last step by some unexplained technique.

 

Might be easier, if you look at it like$$\begin{align*}

df &= f'(\psi) d\psi \\ \\

\int_{f(1)}^{f(0)} df &= \int_1^0 f'(\psi) d\psi \\ \\

f(0) - f(1) &= \int_1^0 f'(\psi) d\psi = \int_1^0 \cos{(\psi^3)} d\psi
\end{align*}$$

 

[Mark44]

From post #1:
$$f(0) = 2 + \int_1^0 \cos(x^3)~ds \approx 2 + (-0.931704) = 1.068296$$
The integral approximation is from wolframalpha - integrate cos(x^3) dx from x=1 to 0 - Wolfram|Alpha

I'm fairly sure that the intent in this problem is that the integral be approximated numerically.

One other thing: why are the integration limits not in the usual order, with the smaller number at the bottom and the larger one at the top?

 

One other thing: why are the integration limits not in the usual order, with the smaller number at the bottom and the larger one at the top?

You can always change it to $2- \int_0^1 f'(\psi) d\psi$ if you want

 

[barryj]

I have attached a file solving a simpler problem two different ways. I learned to solve the problem the first way and this problem uses the second method. I need to get my head around how the second method works. Factchecker, yes the integral was solved using a TI84 calculator. I did not mention this but Mark 44 did.

 

[Mark44]

You can always change it to $2- \int_0^1 f'(\psi) d\psi$ if you want

Yes, I'm well aware of that. My question was why the OP wrote it as shown in post 1.

 

[FranzS]

Homework Statement:: I have this problem and solution. I do not understand the solution, in particular how the integral fits into the solution. Is this a form of the 2nd theorem of calculus perhaps? I am confused.
if f'(x) = cos (x^3) and f(1) = 2 find f(0)
Relevant Equations:: I have attached the problem statement and he solution.

View attachment 273699If f'(x) were a simpler function like f'(x) = cos(x) I would say

f(x) = sin(x) + C and then evaluate C by knowing that 2 = sin(1) + C and then C would equal 2-sin(1)
the f(x) = sin(x) + 2 - sin(1),
f(0) = sin(0) + 2 - sin(1) = 0 + 2 -.841 = 1.58
However the more complicated problem has f'(x) - cos(x^3) and it is solved using an integral expression that is evaluated with a calculator.
I do not understand this integral method. Please help or point me to a link where I can find out what is going on.

They chose the limits of the integral so that both $f(0)$ (goal) and $f(1)$ (known value) pop out as a solution.
In fact, the integral from 1 to 0 gives $f(0)-f(1)$ as a result, so that the solution expression is actually $f(0)=f(1)+f(0)-f(1)$ , which is pretty obvious when written like that.
P.S. You need to solve like that as the indefinite integral $cos(x^3)$ doens't have a closed-form antiderivative (you can't find $f(x)+C$ ). Therefore, a numerical method is applied to solve the definite integral between two limits.

 

[barryj]

My problem with this solution is that I don't understand the "formula" that Mark44 (above) showed. I see what he presented but I don't have a feel for what is going on. I understand the closed form solution but not this numerical solution. What would I ask if I Googled the topic? Numerical solution to an anti derivititive. My calculus books are so old I guess that I have not seen this mentioned. Anyway, I am continuing to try and compare simpler problems to better my understanding.

 

[FactChecker]

Are you asking about the general concept of an exact integral, with the integral end-points known, or about the technique of numerically determining the exact integral? There are several different numerical techniques for integration. That is an entire subject on its own. It is studied within numerical analysis.

 

[barryj]

Actually, I have come to realize that this problems is solved using the Theorem of Calculus. It is really straight forward but I did not see it.