Here's the question you clicked on:
onegirl
Find the function f(x) satisfying the given conditions. (HINT: You are finding the value of C)
Find \[\int\limits_{}^{}f'(x) dx=f(x)+C, \text{ where } (f)'=f'\]
Integrate the function given to get from f' to f.
Don't forget when you integrate, put +C.
\[\int\limits_{}^{}f'(x) dx=\int\limits_{}^{}(4x^2-1) dx\] You may find it easy to use the following: \[\int\limits_{}^{}x^n dx =\frac{x^{n+1}}{n+1}+C, n \neq -1\] This is because \[(\frac{x^{n+1}}{n+1})'=(n+1) \cdot \frac{x^{n+1-1}}{n+1}=x^{n+1-1}=x^n\] And you may also want to use \[\int\limits_{}^{}k dx=kx+C \text{ where} K, C \text{ are a constant} \]
This is because (kx+C)'=(kx)'+(C)'=k(x)'+0+k(x)'=k(1)=k
Are you familiar with the idea of `Integration` yet? Or has it only been introduced to you as the process of finding the `anti-derivative` so far?
@Best_Mathematician can u help?
so, when you have f' (4x^2 +1) take integral to get back the original function. . when take integral, the solution for that method always a family solution. in case they give you the special solution, that means they want you to pick out just special C in set of family solution . take integral of f' you have f(x) = 4x^3/3 -x +C and when they give you f(0) =2 just plug 0 into the function you' ve just got. to find C f(0) = 4 *0^3/3 -0 +C = 2 ---> C =2 that's it
sorry , one more step. plug C back to f(x) = 4x^3/3 - x +2
I 'm supper dummy at teaching, hope you can understand what i mean
so plugging 2 back into f(x) = 4x^3/3 - x + 2 right?
hey, you misunderstand the concept, they ask you to find out C, and you got C =2 that's it. just plug C =2 into the function, nothing to do more.