Sketch the graph and find the area of the region completely enclosed by the graphs of the given functions f and g.

Question

Quizplus · Accepted Answer

The region is enclosed by the graphs of $f(x) = \frac{1}{2}x+2$ and $g(x)=\frac{1}{4}(x-4)^2$.

To find the intersection points, we set them equal:
\begin{align*}
\frac{1}{2}x+2 &= \frac{1}{4}(x-4)^2 \
2x+8 &= (x-4)^2 \
2x+8 &= x^2 - 8x + 16 \
0 &= x^2 - 10x + 8 \
x &= \frac{10\pm\sqrt{84}}{2} = 5\pm\sqrt{21}.
\end{align*}

The parabola $g(x)$ opens upward and its vertex is $(4,0)$. The line $f(x)$ has a positive slope and passes through the point $(0,2)$.

From the intersection points, we see that $5-\sqrt{21} \approx 0.6$ is the left endpoint of the region, and $5+\sqrt{21} \approx 9.4$ is the right endpoint.

Now we can calculate the area:
\begin{align*}
\int_{5-\sqrt{21}}^{5} g(x) \, dx + \int_{5}^{5+\sqrt{21}} f(x) \, dx &= \int_{5-\sqrt{21}}^{5} \frac{1}{4}(x-4)^2 \, dx + \int_{5}^{5+\sqrt{21}} \frac{1}{2}x+2 \, dx \
&= \frac{1}{12}(\sqrt{21}-1)^3 + \frac{1}{4}(5+\sqrt{21}-5)^2 + (\sqrt{21}-1) \
&\approx 4.71.
\end{align*}

Therefore, the best choice is $\boxed{	extbf{(C)}}$.

Sketch the Graph and Find the Area of the Region