Prove Integral Equality of Polynomials Degree 2 & 3

[anemone]

Let $f(x)$ be a polynomial of degree 2 and $g(x)$ a polynomial of degree 3 such that $f(x)=g(x)$ at some three distinct equally spaced points $a,\,\dfrac{a+b}{2}$ and $b$. Prove that $\int_{a}^{b} f(x)\,dx=\int_{a}^{b} g(x)\,dx$.

 

[MarkFL]

My solution:

Spoiler

WLOG, let's orient a new coordinate system such that:

$$f(-c)=g(-c)$$

$$f(0)=g(0)=0$$

$$f(c)=g(c)$$

Hence, we find that:

$$f(x)=u_2x^2+\left(u_1c^2+u_3\right)x$$

$$g(x)=u_1x^3+u_2x^2+u_3x$$

Now, we require:

$$\int_{-c}^0 g(x)-f(x)\,dx=\int_0^c f(x)-g(x)\,dx$$

which, using the FTOC, reduces to:

$$G(-c)-F(-c)=G(c)-F(c)$$

Thus, we require:

$$h(x)=G(x)-F(x)$$

to be an even function. Using $f$ and $g$, we find:

$$G(x)-F(x)=\left(\frac{u_1}{4}x^4+\frac{u_2}{3}x^3+\frac{u_3}{2}x^2\right)-\left(\frac{u_2}{3}x^3+\frac{u_1c^2+u_3}{2}x^2\right)=\frac{u_1}{4}x^4-\frac{u_1c^2}{2}x^2$$

Shown as desired.

 

[anemone]

My solution:

Spoiler

WLOG, let's orient a new coordinate system such that:

$$f(-c)=g(-c)$$

$$f(0)=g(0)=0$$

$$f(c)=g(c)$$

Hence, we find that:

$$f(x)=u_2x^2+\left(u_1c^2+u_3\right)x$$

$$g(x)=u_1x^3+u_2x^2+u_3x$$

Now, we require:

$$\int_{-c}^0 g(x)-f(x)\,dx=\int_0^c f(x)-g(x)\,dx$$

which, using the FTOC, reduces to:

$$G(-c)-F(-c)=G(c)-F(c)$$

Thus, we require:

$$h(x)=G(x)-F(x)$$

to be an even function. Using $f$ and $g$, we find:

$$G(x)-F(x)=\left(\frac{u_1}{4}x^4+\frac{u_2}{3}x^3+\frac{u_3}{2}x^2\right)-\left(\frac{u_2}{3}x^3+\frac{u_1c^2+u_3}{2}x^2\right)=\frac{u_1}{4}x^4-\frac{u_1c^2}{2}x^2$$

Shown as desired.

Well done, MarkFL! Your solution is equally elegant as the one I have at hand and since both solutions differ a little, I'll hence share it here:

Spoiler

For simplicity, we translate the origin to the midpoint of the interval $(a,\,b)$.

Let $k=\dfrac{b-a}{2}$, $F(x)=f\left(x+\dfrac{a+b}{2} \right)$ and $G(x)=g\left(x+\dfrac{a+b}{2} \right)$ then we have

1. $F(-k)=f(a)=g(a)=G(-k)$

2. $F(0)=f\left(\dfrac{a+b}{2} \right)=g\left(\dfrac{a+b}{2} \right)=G(0)$

3. $F(k)=f(b)=g(b)=G(k)$.

Also, $\int_{a}^{b} f(x)\,dx=\int_{-k}^{k} F(x)\,dx$ and $\int_{a}^{b} g(x)\,dx=\int_{-k}^{k} G(x)\,dx$.

So it suffices to show that $\int_{-k}^{k} F(x)\,dx=\int_{-k}^{k} G(x)\,dx$.

We then let $P(x)=F(x)-G(x)$. Then $P(x)$ is a polynomial of degree 3 with zeros at $-k,\,0$ and $k$ so for some constant $c$,

$P(x)=cx(x-k)(x+k)=cx(x^2-k^2)$.

From the above, we see that $P$ is an odd function, and therefore $\int_{-k}^{k} P(x)\,dx=0$ and we're done.