Hello, Gracy!

Graph of $\displaystyle f(x)$ consists of line segments and semicircles.

Use it to geometrically evaluate the following integrals.

$\displaystyle (1)\;\int^{14}_0\!\!f(x)\,dx \qquad (2)\;\int^{10}_0\!\!f(x)\,dx \qquad (3)\;\int^{12}_3\!\!f(x)\,dx$ Code:

* *
* *
* 9π/2 *
* *
*-------------*
| | *
| 12 | *
| | 2 * 10 14
--+------+------+-------*-------+---------+
0 6 8 * -2 | |
* | -8 |
* | |
*---------*
* -2π *
* *
* *

At the far left, there is a semicircle of radius 3.

. . Its area is: .$\displaystyle \frac{1}{2}\pi(3^2) \:=\:\frac{9\pi}{2}$

Below it is a 6-by-2 rectangle; its area is $\displaystyle 12$

There are two isoceles right triangles, each with area $\displaystyle 2.$

At the far left, there is a semicircle of radius 2.

. . Its area is: .$\displaystyle \frac{1}{2}\pi(2^2) \:=\:2\pi$

Above it is a 4-by-2 rectangle; its area is $\displaystyle 8.$

The regions below the x-axis have "negative areas".

$\displaystyle (1)\;\int^{14}_9f(x)\,dx\;=\;\frac{9\pi}{2} + 12 + 2 - 2 - 2\pi^2 - 8 \;=\;\frac{\pi}{4} + 4$

$\displaystyle (2)\;\int^{10}_0f(x)\,dx\;=\;\frac{9\pi}{2} + 12 + 2 - 2 \;=\;\frac{9\pi}{2} + 12$

$\displaystyle (3)\;\int^{12}_3f(x)\,dx\;=\;\frac{1}{2}\left(\fra c{9\pi}{2} + 12\right) + 2 - 2 - \frac{1}{2}\left(2\pi + 8\right) \;= \;\frac{5\pi}{4} + 2$