SOLUTION: f (x) is polynomial function, f '(x) + int f (x) dx = x ^4 + 13 x ^2 + 2. Find f (x)

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Question 1207322: f (x) is polynomial function, f '(x) + int f (x) dx = x ^4 + 13 x ^2 + 2. Find f (x)
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!




int%28%22f%28x%29%22%2Cdx%29%22%22=%22%22

Since the sum of those must be identically equal to x%5E4%2B13x%5E2%2B2,

then the largest power of x that occurs in the sum, which is n+1, must be
the largest power that occurs in x%5E4%2B13x%5E2%2B2, which is 4. Therefore,
n+1 = 4 and n=3.  Substituting (and eliminating the 3 dots since they
are no longer needed):

%22f%28x%29%22=a%5B0%5Dx%5E3%2Ba%5B1%5Dx%5E%283-1%29%2Ba%5B3-1%5Dx%2Ba%5B3%5D
%22f%27%28x%29%22=3%2Aa%5B0%5Dx%5E%283-1%29%2B%283-1%29%2Aa%5B1%5Dx%5E%283-2%29%2Ba%5B3-1%5D
int%28%22f%28x%29%22%2Cdx%29%22%22=%22%22

Simplifying,

%22f%28x%29%22=a%5B0%5Dx%5E3%2Ba%5B1%5Dx%5E2%2Ba%5B2%5Dx%2Ba%5B3%5D
%22f%27%28x%29%22=3%2Aa%5B0%5Dx%5E2%2B2%2Aa%5B1%5Dx%2Ba%5B2%5D
int%28%22f%28x%29%22%2Cdx%29%22%22=%22%22

Since the coefficients of x3 and x are 0 in x%5E4%2B13x%5E2%2B2,

a%5B0%5Dx%5E3%2Bexpr%281%2F3%29%2Aa%5B1%5Dx%5E3=0 and a%5B2%5Dx%2B2a%5B1%5Dx%2Ba%5B3%5D=0.

Since the coefficient of x2 is 13, a%5B1%5Dx%5E2%2B3a%5B0%5Dx%5E2%2Bexpr%281%2F2%29a%5B2%5Dx%5E2=13 

and since its constant term is 2, a%5B3%5D%2Ba%5B2%5D%2BC=2

So we have the system of equations



Your teacher must have meant for you to take the constant 
of integration to be 0.  Otherwise you have 5 unknowns but 
only 4 equations.  I'll use my system solver to solve that.  
I get: 

system%28a%280%29=-1%2F3%2Ca%281%29=1%2Ca%282%29=26%2Ca%283%29=-26%29

So that makes

%22f%28x%29%22=expr%28-1%2F3%29x%5E3%2Bx%5E2%2B26x-26

Edwin

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
f (x) is  highlight%28highlight%28a%29%29  polynomial function, f '(x) + int f (x) dx = x ^4 + 13 x ^2 + 2,
Find f(x)
~~~~~~~~~~~~~~~~~~~~~~

We want to find f(x) as a polynomial  f(x) = a%5B0%5D%2Ax%5En + a%5B1%5D%2Ax%5E%28n-1%29 + . . . + a%5Bn%5D.

Taking derivative decreases the degree of a polynomial by one unit.
Taking antiderivative increases the degree of a polynomial by one unit.


Since the sum   f ' (x) + int f (x) dx   is a polynomial of degree 4,
                ----------------------

it means that the sough polynomial f(x) is of degree 3:

    f(x) = ax^3 + bx^2 + cx + d.


Then 

    f ' (x)     =                        3a%2Ax%5E2 + 2bx + c,

    int f(x) dx =  %28a%2F4%29x%5E4 + %28b%2F3%29x%5E3 + %28c%2F2%29x%5E2 + dx  + E.



So, in the sum   f ' (x) + int f(x) dx
                 ----------------------

    (a)  coefficient at  x%5E4  is  a%2F4  It gives an equation

             a%2F4 = 1;   hence  a = 4.



    (b)  coefficient at  x%5E3  is  0.  It gives an equation

             b%2F3 = 0;   hence  b = 0.



    (c)  coefficient at  x%5E2  is  13.  It gives an equation

             3a%2B+c%2F2 = 13,  or  3%2A4+%2B+c%2F2 = 13  ---> c%2F2 = 13 - 12 = 1  --->  c = 2.



    (d)  coefficient at  x  is  0.  It gives an equation

             2b + d = 0,  which implies  2*0 + d = 0;  hence,  d = 0.



    +------------------------------------------------------------+
    |    At this point, the problem is just solved to the end.   |
    |            a = 4;  b = 0;  c = 2;  d = 0.                  |
    +------------------------------------------------------------+



The sough polynomial is  f(x) = 4x^3 + 2x.     ANSWER



CHECK.  The derivative is f ' (x) = 12x%5E2+%2B+2.

        The anti-derivative is  F(x) =  %284%2F4%29x%5E4+%2B+%282%2F2%29x%5E2 = x%5E4+%2B+x%5E2.

        The sum f ' (x) + F(x) = %2812x%5E2%2B2%29 + %28x%5E4+%2B+x%5E2%29 = x%5E4+%2B+13x%5E2+%2B+2.   ! correct !

Solved.

Do not accept any other  highlight%28different%29  answer.


--------------


The solution by  Edwin is  INCORRECT.

To make sure that it is incorrect,  simply take the antiderivative of his leading term  %28-1%2F3%29x%5E3.

This antiderivative is  %28-1%2F%283%2A4%29%29%2Ax%5E4 = %28-1%2F12%29x%5E4,  and no other arguments are needed anymore.