You are given the four points in the plane a=(-2,2), b=(1,-3), c=(3,6), and d=(7,-3). The graph of the function consists of the three line segments AB, BC and CD. Find the integral by interpreting the integral in terms of sums and/or differences of areas of elementary figures

Well, I gotta say, these points really know how to keep things interesting! It's like they're playing a game of connect the dots. A math-themed connect the dots, if you will.

So, let's start by drawing the graph. Picture this: you have point A at (-2,2), point B at (1,-3), point C at (3,6), and point D at (7,-3). And you want to connect them with line segments AB, BC, and CD.

Now, let's talk about integrating. It's like the superpower of calculus, where you're finding the area under a curve. But in this case, we're dealing with line segments, so things are a little different.

To find the integral , we need to interpret it in terms of sums and differences of areas of elementary figures. That means we'll break the shapes formed by the line segments into smaller, more manageable pieces.

First, let's look at the line segment AB. It's just a straight line connecting A and B. Since it's a line, we know it's easy to find the area under it, right? Well, unfortunately, lines don't really have area. They're like the skinny jeans of geometry. So, for AB, our area is zero.

Next, we move on to BC. This time, we have a diagonal line connecting B and C. Again, lines don't have area, so we're out of luck there. Another zero for us.

Finally, we have CD, another straight line segment. And once again, lines don't have area. So, we're batting zero for three here.

Therefore, when we add up the areas of these elementary figures, we get a big fat zero as our integral! Now that's what I call a summation of nothingness. It's like winning the award for the best paperclip collection – not exactly impressive, but still an achievement in its own way.

I hope my clownish explanation brought a smile to your face while addressing your question! If you need any further clarification, feel free to ask!

To find the integral of the function, we first need to find the equation of the three line segments and then calculate the area under the curve.

1. Line Segment AB:
The equation of the line passing through points A and B can be found using the slope-intercept form (y = mx + b).
The slope (m) can be calculated as (change in y)/(change in x) = (-3 - 2)/(1 - (-2)) = -5/3.
Using point A(-2,2), we can substitute the values in the slope-intercept form and solve for b:
2 = (-5/3)(-2) + b
2 = 10/3 + b
b = 2 - 10/3
b = -4/3

Thus, the equation of line AB is y = (-5/3)x - 4/3.

2. Line Segment BC:
The equation of the line passing through points B and C can be found similarly.
The slope (m) can be calculated as (change in y)/(change in x) = (6 - (-3))/(3 - 1) = 9/2.
Using point B(1,-3), we can substitute the values in the slope-intercept form and solve for b:
-3 = (9/2)(1) + b
-3 = 9/2 + b
b = -3 - 9/2
b = -15/2

Thus, the equation of line BC is y = (9/2)x - 15/2.

3. Line Segment CD:
The equation of the line passing through points C and D can be found similarly.
The slope (m) can be calculated as (change in y)/(change in x) = (-3 - 6)/(7 - 3) = -3.
Using point C(3,6), we can substitute the values in the slope-intercept form and solve for b:
6 = (-3)(3) + b
6 = -9 + b
b = 6 + 9
b = 15

Thus, the equation of line CD is y = -3x + 15.

Now, we need to find the x-coordinates of the points where the line segments intersect:
- Set the equations of AB and BC equal to each other:
(-5/3)x - 4/3 = (9/2)x - 15/2
Multiply through by 6:
-10x - 8 = 27x - 45
Combine like terms:
37x = 37
x = 1

- Set the equations of BC and CD equal to each other:
(9/2)x - 15/2 = -3x + 15
Multiply through by 2:
9x - 15 = -6x + 30
Combine like terms:
15x = 45
x = 3

Now, we calculate the area under the curve by interpreting the integral as the difference of areas of elementary figures.

The integral of the function from x = 1 to x = 3 is given by:
∫(1 to 3) [(-5/3)x - 4/3 - ((9/2)x - 15/2)] dx + ∫(3 to 7) [-3x + 15 - ((9/2)x - 15/2)] dx

Simplifying the integral, we get:
= ∫(1 to 3) [(4/3) - (4/3)x] dx + ∫(3 to 7) [(3/2)x] dx

Integrating the individual terms, we get:
= [(4/3)x - (2/3)x^2] |(1 to 3) + [(3/4)x^2] |(3 to 7)

= [(4/3)(3) - (2/3)(3)^2] - [(4/3)(1) - (2/3)(1)^2] + [(3/4)(7)^2] - [(3/4)(3)^2]

Simplifying further, we get:
= [4 - 6] - [4/3 - 2/3] + [63/4] - [9/4]

= -2 - 2/3 + 63/4 - 9/4

= -2/3 + 63/4 - 9/4

= -2/3 + 54/4

= -2/3 + 27/2

Now, we can simplify the fraction further:
= [-4/6 + 81/6] / 3

= 77/6 / 3

= 77/6 * 1/3

= 77/18

Therefore, the integral from x = 1 to x = 3 is equal to 77/18.

To find the integral of the function f(x) over the graph of the piecewise continuous function, we need to interpret it in terms of sums and/or differences of areas of elementary figures.

Let's break down the problem into three separate line segments: AB, BC, and CD.

1. For the line segment AB:
We need to find the area under the curve f(x) between the x-coordinates of point A (-2) and point B (1). To do this, we can calculate the area of the trapezoid formed by the line segment AB and the x-axis.

The equation of the line passing through A and B can be determined using the slope formula:
slope = (y2 - y1) / (x2 - x1)
slope = (-3 - 2) / (1 - (-2))
slope = -5/3

The equation of the line passing through A(-2, 2) with a slope of -5/3 can be written as:
y = (-5/3)(x + 2) + 2

Now, we can calculate the areas.

First, find the x-coordinate of the intersection point between the line AB and the x-axis by setting y = 0:
0 = (-5/3)(x + 2) + 2
0 = (-5/3)x - (10/3)
(5/3)x = -10/3
x = -2

The area of the trapezoid AB can be calculated as follows:
Area_AB = 0.5 * (base1 + base2) * height
= 0.5 * (3 + 1) * |-2 - (-2)|
= 0.5 * 4 * 0
= 0

2. For the line segment BC:
We need to find the area under the constant function f(x) = -3 between the x-coordinates of point B (1) and point C (3), which is simply a rectangle.

The width of the rectangle is given by the difference in x-coordinates: 3 - 1 = 2.
The height of the rectangle is the value of f(x) = -3.

Area_BC = width * height
= 2 * (-3)
= -6

3. For the line segment CD:
We need to find the area under the curve f(x) between the x-coordinates of point C (3) and point D (7).

First, calculate the equation of the line passing through C and D. We use the slope formula again:
slope = (y2 - y1) / (x2 - x1)
slope = (-3 - 6) / (7 - 3)
slope = -9/4

The equation of the line passing through C(3, 6) with a slope of -9/4 can be written as:
y = (-9/4)(x - 3) + 6

Now, we can calculate the areas.

First, find the x-coordinate of the intersection point between the line CD and the x-axis by setting y = 0:
0 = (-9/4)(x - 3) + 6
0 = (-9/4)x + (27/4) + 6
(9/4)x = (27/4) + 6
(9/4)x = (27/4) + (24/4)
(9/4)x = (51/4)
x = 51/36
x = 17/12

The area of the trapezoid CD can be calculated as follows:
Area_CD = 0.5 * (base1 + base2) * height
= 0.5 * (7 + 17/12 - 3 - 17/12) * |(17/12) - 3|
= 0.5 * (74/12) * (5/12)
= (74/24) * (5/12)
= (5 * 74) / (24 * 12)
= 185/288

Finally, we can calculate the integral by summing up the areas of the elementary figures:
∫[a,d]f(x) dx = Area_AB + Area_BC + Area_CD
= 0 + (-6) + 185/288
= -6 + 185/288
= (288 * (-6) + 185) / 288
= (-1728 + 185) / 288
= -1543 / 288

Therefore, the integral of the function f(x) over the given graph is -1543/288.

Here the integral is being evaluated, so the area of the elementary figures could be positive or negative.

It can be observed that all the line segments form a (twisted) trapezoid with the x-axis, i.e. part of the area is above the x-axis, and part of it is below.

Knowing that the area of a trapezoid is the average of the y-coordinate of the two ends, multiplied by the length, it will be possible to calculate the three areas separately.

For example, the area created by the segment AB is:
Average of y-coordinates = (2+(-3))/2=-0.5
Length = (1-(-2)) =3
Integral of segment AB=3*(-0.5) = -1.5

Repeat for the two other segments and add up the total.