Find the area of the region bounded by the graphs of the algebraic functions.

Question

Quizplus · Accepted Answer

To find the area, we need to find the points of intersection of the graphs. 
$$\begin{cases} y = x^3 - 4x^2 + 5x \ y = x^2 - 3x + 2 \end{cases}$$
Setting them equal to each other, 
$$x^3 - 4x^2 + 5x = x^2 - 3x + 2$$
$$x^3 - 5x^2 + 8x - 2 = 0$$
We can use synthetic division or rational root theorem to find one root as 2. Dividing by $x - 2$ gives us
$$x^2 - 3x +1$$
Using quadratic formula, the other two roots are $\dfrac{3 \pm \sqrt{5}}{2}$
Now we can integrate to find the area:
$$\int_0^{\frac{3-\sqrt{5}}{2}} (x^2 - 3x +2) dx + \int_{\frac{3-\sqrt{5}}{2}}^2 (x^3 - 4x^2 +5x - x^2 +3x - 2) dx + \int_2^{\frac{3+\sqrt{5}}{2}} (x^2 - 3x +2) dx$$
Simplifying,
$$\int_0^{\frac{3-\sqrt{5}}{2}} (x^2 - 3x +2) dx + \int_{\frac{3-\sqrt{5}}{2}}^2 (x^3 - 5x^2 +2x - 2) dx + \int_2^{\frac{3+\sqrt{5}}{2}} (x^2 - 3x +2) dx$$
Evaluating the integrals gives us the final answer $\dfrac{8 - \sqrt{5}}{6} + \dfrac{7-2\sqrt{5}}{6} + \dfrac{8 - \sqrt{5}}{6} = \dfrac{17-4\sqrt{5}}{3}$, which is approximately 1.407. The region is bounded by the graphs of $y=2x+2$ and $y=-\frac{1}{2}x+\frac{7}{2}$.

Setting these two equations equal gives $2x+2=-\frac{1}{2}x+\frac{7}{2}$, which simplifies to $x=1$. Thus, we can find the $y$-values of the intersection by plugging in $x=1$: $y_{1}=2\cdot 1+2=4$ and $y_{2}=-\frac{1}{2}\cdot 1+\frac{7}{2}=\frac{5}{2}$.

The intersection point is $(1,\frac{5}{2})$.

Now we need to integrate the two functions with respect to $x$ to find the area:
\begin{align*}
A &= \int_{0}^{1} (2x+2) dx - \int_{0}^{1} \left(-\frac{1}{2}x+\frac{7}{2}ight) dx \
&= \left[x^{2}+2xight]_{0}^{1} - \left[-\frac{1}{4}x^{2}+\frac{7}{2}xight]_{0}^{1} \
&= \left(1+2ight)-\left(-\frac{1}{4}+\frac{7}{2}ight) \
&= \boxed{\frac{15}{4}}
\end{align*}

Find the Area of the Region Bounded by the Graphs