# area under graphs using integration

• Jul 26th 2009, 11:45 PM
Mr.Ree
area under graphs using integration
Hello. I am new to this forum, and starting a new topic in calculus. I need some help with integration to find areas under certain curves (see picture).

I worked out the two trig functions to be:

1) cos ((pi*x)/2)]

because: positive cosine function
Amplitude = 1
Period = 4 = [2*pi]/B
-> B = 2*pi/4
-> B = pi/2

2) -cos ((pi*x)/2)

because: negative cosine function
Amplitude = 1
Period = 4 = [2*pi]/B
-> B = 2*pi/4
-> B = pi/2

How do I integrate and find the areas under these graphs?

The other 4 functions which look like exponentials I need some help with:

I want to use: y = A*e^(kx) and manipulations of that, but I am wondering whether substituting my points will work because I have no initial value since y intercept is not given. How can I find the equations of the exponential functions from just the graph. (I'd love to learn how without a calculator)
• Jul 27th 2009, 08:27 AM
skeeter
Quote:

Originally Posted by Mr.Ree
Hello. I am new to this forum, and starting a new topic in calculus. I need some help with integration to find areas under certain curves (see picture).

I worked out the two trig functions to be:

1) cos ((pi*x)/2)]

because: positive cosine function
Amplitude = 1
Period = 4 = [2*pi]/B
-> B = 2*pi/4
-> B = pi/2

2) -cos ((pi*x)/2)

because: negative cosine function
Amplitude = 1
Period = 4 = [2*pi]/B
-> B = 2*pi/4
-> B = pi/2

How do I integrate and find the areas under these graphs?

The other 4 functions which look like exponentials I need some help with:

I want to use: y = A*e^(kx) and manipulations of that, but I am wondering whether substituting my points will work because I have no initial value since y intercept is not given. How can I find the equations of the exponential functions from just the graph. (I'd love to learn how without a calculator)

the area of one "arch" of the cosine graph is ...

$\int_{-1}^1 \cos\left(\frac{\pi x}{2}\right) \, dx = 2\int_0^1 \cos\left(\frac{\pi x}{2}\right) \, dx$

so, the area for a single pair of "arches" between the two opposite cosine graphs would be ...

$4\int_0^1 \cos\left(\frac{\pi x}{2}\right) \, dx$

$4\left[\frac{2\sin\left(\frac{\pi x}{2}\right)}{\pi}\right]_0^1$

$4\left[\frac{2}{\pi} - 0\right] = \frac{8}{\pi}$

as far as your "exponential" curves, $y = Ae^{kx}$ , $A$ is the y-intercept for each curve in each quadrant, so you do need that piece of info. You will need another point on each of the four curves to determine correct values for $k$.