Question

(a) Derive the difference equation corresponding to the approximation of integration found by fitting a parabolas to the points ek-2, ek-1, ek and taking the area under this parabola between t = kT –T and t = kT as the approximation to the integral of e(t) over this interval. (b) Find the transfer function of the resulting discrete system, and plot the poles and zeros in the Z – plane

Answer 1

Integrate from [-2, 2].

Step 1: Set up the integral.

Step 2: Find the Integral.

*Note: We don't have to add a "+C" at the end because it will cancel out finding the area anyway.

Step 3: Integrate from the given interval, [-2,2].

The area of the curve to the x axis from -2 to 2 is ³²⁄₃ units squared.

On the graph, the red below the parabola is the area and the dotted line is the integral function. Notice that the integral function is cubic and the original function is quadratic. The integral will always be a degree higher than the original function. Looking at the graph, there is a geometric relationship between the original function and the integral function. We can see at x = -2 the integral function has a y value of a little under -5, and at x = 2 the integral has a y value of a little over 5. The difference of 5.3 and -5.3 gives us an area of ³²⁄₃, which is a little over 10.

When taking the definite integral over an interval, sometimes we will get negative area because the graph interprets area above the x axis as positive area and below the x axis as negative area.

Find the Area with Integration Examples

(2) Let's take the integral of y = x from [-3, 1].

We end up with an area of -4.

Looking at the graph at x = -3 and 1, the integral function has an F(x) of 4.5 and 0.5 respectively. Subtracting the lower bound value (4.5) from the upper bound value (0.5) will yield -4. If we wanted to find the total area, we could take the absolute value of each bound and add them together to get 5.

It is possible to integrate a function that is not continuous, but sometimes we need to break up the area into two different integrals.

(3) Here is the function.

Before integrating, we should graph this function to see what it looks like.

Evaluate each integrand.

a)

b)

a) Since the original function is not continuous, we need to look at the bounds first to see if we are integrating through any discontinuous points. Since [3,10] is greater than 1, it is continuous and we can integrate using one integral.

b) This integral is a little different. The interval is discontinuous from [-2,2], so we need to split it into two integrals and add them together.

We have one integral with the interval [-2,1] and the other from [1,2]. The first interval is less than 1 and the second is greater than 1.

After we integrate, we plug in the specific bounds for both.

The area from [-2,2] is 14 units squared and the area from [3,10] is 35 units squared.

(4) Find the general integral for the yellow shaded region

The area is the integral of f minus the area of g.

(5) Find the area of the purple region bounded by three lines:

First, we need to find the three points of intersection to establish our intervals for integration. We set each function equal and solve for x.

Like (4), we have to subtract integrals, but we have two seperate quantities to add.

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