Getting the area between $y$-axis, $f(x)$, $f(2-x)$ when both function is given by their subtraction?

70 Views Asked by At
  • For a polynomial $f(x)$, let function $g(x) = f(x) - f(2-x)$.
  • $g'(x) = 24x^2 - 48x + 50$

What is the area between $y=f(x)$, $y=f(2-x)$, and $y$-axis?

My approach:

  1. $g(1) = f(1) - f(1) = 0$.
  2. From $g'(x)$, $g(x) = 8x^3 - 24x^2 + 50x + C$.
  3. From 1 and 2, $C = -34$.
  4. Since $f(x)$ is a cubic function, let $f(x) = ax^3 + bx^2 + cx + d$ and compare coefficients of $g$ and $f(x) - f(2-x)$.
  5. $a = 4$, $2b + c = 1$.

And I'm stuck. I can't see how I can get more info about $f(x)$ using these conditions and get areas out of it.

1

There are 1 best solutions below

2
On BEST ANSWER

From where you left off, $$ g(x) \ \ = \ \ f(x) \ - \ f(2 - x) $$ $$ = \ \ 2a·x^3 \ - \ 6a·x^2 \ + \ (12a + 4b + 2c)·x \ - \ (8a + 4b + 2c) \ \ , $$

which does match up with your integration of $ \ g'(x) \ $ with $ \ a \ = \ 4 \ \ , \ \ 2b + c \ = \ 1 \ \ : $

$$ g(x) \ \ = \ \ 2·4·x^3 \ - \ 6·4·x^2 \ + \ (12·4 + 2·1)·x \ - \ (8·4 + 2·1) $$ $$ = \ \ 8·x^3 \ - \ 24·x^2 \ + \ 50·x \ - \ 34 \ \ . $$

We would need to express $ \ f(x) \ $ as, say, $ \ 4x^3 + bx^2 + (1 - 2b)·x + d \ \ , \ $ giving us $$ \ f(2 - x) \ = \ -4x^3 \ + \ (b + 24)·x^2 \ - \ (2b + 49)·x \ + \ (d + 34) \ \ \ . $$

This tells us that the $ \ y-$intercept of $ \ f(2 - x) \ $ is above that of $ \ f(x) \ \ . \ $ Since $ \ f(2 - x) \ $ is the vertical reflection of $ \ f(x) \ $ about the $ \ y-$axis, "shifted to the right" by 2 units, it is also the vertical reflection of $ \ f(x) \ $ about the line $ \ x \ = \ 1 \ \ . \ $ So the two function curves intersect at $ \ x \ = \ 1 \ \ $ [the geometrical meaning of your result $ \ g(1) \ = \ 0 \ ] \ $ with $ \ f(2 - x) \ $ "above" $ \ f(x) \ \ . $ [We know that there are no other intersections in the interval $ \ ( 0 \ , \ 1 ) \ $ because $ \ g(0) \ = \ -34 \ $ and $ \ g'(x) \ = \ 24·(x - 1)^2 + 26 \ > \ 0 \ \ . \ ] $

For the problem then, we wish to integrate $$ \ \int_0^1 \ [ \ f(2 - x) \ - \ f(x) \ ] \ \ dx \ \ = \ \ \int_0^1 \ -g(x) \ \ dx \ \ . $$

[The value of $ \ d \ $ would be irrelevant in any case for the area between the curves and we find that we also don't need to know $ \ b \ \ . \ $ We do obtain a positive value for this area between the curves, as we should.]