1. ## integral subdivision rule

Graph of f consistes of linesegments and semicircles.Use it to geometrically evaluate the following integrals.
(1) integral (0 to 14) f(x) dx

(2) integral (0 to 10) f(x) dx
(3) integral (3 to 12) f(x) dx

For diagram please see the doc fileattached.

2. Originally Posted by gracy
Graph of f consistes of linesegments and semicircles.Use it to geometrically evaluate the following integrals.
(1) integral (0 to 14) f(x) dx

(2) integral (0 to 10) f(x) dx
(3) integral (3 to 12) f(x) dx

For diagram please see the doc fileattached.
Integral means you have to find the area between graph and x axis.
If graph is above x-axis, tha area is positive
if graph is below x axis, area is negative
integral (0 to 6) f(x) dx=6*2 +area of semicicle= $12+\pi *3^2*\frac{1}{2}$= $12+4.5\pi$
integral (6 to 8) f(x) dx=area of triangle of base length 2 and heght 2=2
integral (8 to 10) f(x) dx=-2
integral (0 to 6) f(x) dx=-(4*2 +area of semicicle)= $-(8+\pi *2^2*\frac{1}{2})$= $-(8+2\pi)$

1) $4+2.5\pi$
2) $12+4.5\pi$
for 3) you have to find the area between 3 and 6
It will be equal to half of area between 0 and 6 doue to symmetry.
3) $-2-2.5\pi$

Keep Smiling
Malay

3. Hello, Gracy!

Graph of $f(x)$ consists of line segments and semicircles.
Use it to geometrically evaluate the following integrals.

$(1)\;\int^{14}_0\!\!f(x)\,dx \qquad (2)\;\int^{10}_0\!\!f(x)\,dx \qquad (3)\;\int^{12}_3\!\!f(x)\,dx$
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: . $\frac{1}{2}\pi(3^2) \:=\:\frac{9\pi}{2}$
Below it is a 6-by-2 rectangle; its area is $12$

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

At the far left, there is a semicircle of radius 2.
. . Its area is: . $\frac{1}{2}\pi(2^2) \:=\:2\pi$
Above it is a 4-by-2 rectangle; its area is $8.$

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

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

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

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