# Help with this Calculus Problem

• May 30th 2006, 01:42 PM
Susie38
Help with this Calculus Problem
This is a piecewise function
Let f(x) = 2/pi*x when [0,pi/2]
2sinx when (pi/2, 3pi/2]
1 when (3pi/2, 2pi]

a. find a continuous function, F(x), satisfying F(x) = the integral f(t) dt from 0 to x, on [0,2pi]

b. Give the values of x on [0,2pi] for which F(x) is not differentiable.

c. compute the integral f(x) dx from 0 to 2pi.
• May 30th 2006, 06:03 PM
Soroban
Hello, Susie38!

Here are parts (b) and (c).

Quote:

$f(x) \:= \:\begin{Bmatrix}\frac{2}{\pi}x & 0 \leq x \leq \frac{\pi}{2} \\ 2\sin x & \frac{\pi}{2} < x \leq \frac{3\pi}{2} \\ 1 & \frac{3\pi}{2} < x \leq 2\pi\end{Bmatrix}$

a. Find a continuous function, $F(x)$, satisfying $F(x) = \int^x_0 f(t)\,dt$

b. Give the values of $x$ on $[0,2\pi]$ for which $F(x)$ is not differentiable.

c. Compute: $\int^{2\pi}_0 f(x)\,dx$
On $\left[0,\,\frac{\pi}{2}\right]\]$, the function is a straight line.

On $\left(\frac{\pi}{2},\,\frac{3\pi}{2}\right]$, the function is a sine curve.

On $\left(\frac{3\pi}{2},\,2\pi\right]$, the function is a horizontal line.
Code:

        |       1 +      **              o * * * *         |    * :  *          :      :         |  *  :    *        :      :         | *    :      *        :      :       - * - - - + - - - * - - - + - - - + -         |      π/2    2π*    3π/2    2π         |                *    :         |                  *  :     -1 +                      **         |
(b) $F(x)$ is not differentiable at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$

It has two different slopes at $x = \frac{\pi}{2}$

It is discontinuous at $x = \frac{3\pi}{2}$

(c) No calculus needed!

The triangle has base $\frac{\pi}{2}$ and height $1.$
. . . Its area is: $A_1 \:= \:\frac{1}{2}\left(\frac{\pi}{2}\right)(1)\:=\; \frac{\pi}{4}$

The sine curve has a region above and a region below the x-axis.
. . . The two regions have equal area and will cancel out.
Its net area is: $A_2\:=\:0$

The rectangle has base $\frac{\pi}{2}$ and height $1.$
. . . Its area is: $A_3\:=\:\left(\frac{\pi}{2}\right)(1)\:=\: \frac{\pi}{2}$

The value of the integral is the sum of these areas:
. . . $A_1 + A_2 + A_3 \;= \;\frac{\pi}{4} + 0 + \frac{\pi}{2}\;=\;\frac{3\pi}{4}$