Area enclosed by two curves

Suppose the $x$ -coordinates of the two intersection point are $d$ and $e$ , where $e>d$ . Then it is possible to write $\displaystyle g(x)-f(x)=x^4 + ax^3 + bx +1=(x-d)^2(x-e)^2$ .

Then the the desired area, $\displaystyle A$ is $\int_d^e \left(g(x)-f(x)\right)\, dx=\int_d^e (x-d)^2(x-e)^2\, dx$

Let $\displaystyle r=\frac{e-d}{2}$ and $\displaystyle u=x-\frac{d+e}{2}$ . Then the area is

$A= \int_{-r}^r(u-r)^2(u+r)^2\, du =\int_{-r}^r (u^2-r^2)^2\, du=2\int_{0}^r (u^2-r^2)^2\, du=\ldots = \frac{16}{15}r^5$

In order to obtain the value of $r$ , By comparing each of the coefficient of $x^k$ of $\displaystyle f-g$ respectively, one can obtain

$\begin{aligned} -2(d+e) &=& a\\ d^2+4de+e^2 &=& 0 \\ -2de(d+e) &=&b\\ d^2e^2&=&1 \end{aligned}$

From the last equation, $\displaystyle de= \pm 1$ . If $\displaystyle de= 1$ , then from the second equation, $\displaystyle d^2+4de+e^2=0$ , a contradiction. Hence, $\displaystyle de= -1$ . From second equation again, $\displaystyle d^2+e^2=-4de=4$ , which means that $\displaystyle (d-e)^2=4-2de=6$ . As $\displaystyle e>d, e-d=\sqrt{6}$ . Now, $\displaystyle r=\frac{e-d}{2} =\frac{\sqrt{6}}{2}$ and hence $\displaystyle A= \frac{16}{15}\times \frac{36 \sqrt{6}}{32}=\frac{6\sqrt{6}}{5}$ .

Bonus : If the parabola $f(x)=x^2$ is tangent to the graph of $g(x) = x^4 + ax^3 + \color{#D61F06}cx^2+ bx +1$ at two distinct points, what is the area by $f$ and $g$ ?

The answer is $\displaystyle \frac{(\color{#D61F06}c+5)^2\sqrt{\color{#D61F06}c+5}}{30}$