You are given the four points in the plane A(-3,6),B(0,-6),C(5,7),D(8/-7) . The graph of the function f(x) consists of the three line segments AB ,BC and CD . Find the integral f(x)from -3 to by interpreting the integral in terms of sums and/or differences of areas of elementary figures.
To find the integral of f(x) from -3 to b, we need to calculate the area between the graph of f(x) and the x-axis within the given interval.
Let's start by finding the equation of each line segment.
1. Line segment AB:
Using the coordinates A(-3,6) and B(0,-6), we can find the equation of the line passing through these points using the slope-intercept form (y = mx + b).
The slope (m) can be calculated as (change in y)/(change in x) = (6 - (-6))/(-3 - 0) = 12/-3 = -4.
Substituting the coordinates of point A into the equation, we get:
6 = -4(-3) + b
6 = 12 + b
b = -6
Therefore, the equation of line AB is y = -4x - 6.
2. Line segment BC:
Using the coordinates B(0,-6) and C(5,7), we can find the equation of the line passing through these points using the slope-intercept form.
The slope (m) can be calculated as (change in y)/(change in x) = (7 - (-6))/(5 - 0) = 13/5.
Substituting the coordinates of point B into the equation, we get:
-6 = (13/5)(0) + b
-6 = b
Therefore, the equation of line BC is y = (13/5)x - 6.
3. Line segment CD:
Using the coordinates C(5,7) and D(8,-7), we can find the equation of the line passing through these points using the slope-intercept form.
The slope (m) can be calculated as (change in y)/(change in x) = (-7 - 7)/(8 - 5) = -14/3.
Substituting the coordinates of point C into the equation, we get:
7 = (-14/3)(5) + b
7 = -70/3 + b
b = 91/3
Therefore, the equation of line CD is y = (-14/3)x + 91/3.
Now, we can find the integral of f(x) from -3 to b by calculating the areas between the graph of f(x) and the x-axis within the given interval.
To do this, we divide the interval -3 to b into three subintervals: -3 to 0, 0 to 5, and 5 to b.
1. Area of the region under the line segment AB within the interval -3 to 0:
The equation of line AB is y = -4x - 6. To find the area under this line segment within the interval -3 to 0, we calculate the area of the corresponding trapezoid.
The base lengths are 3 units (the x-distance) and the heights are 6 units (the y-values at x=-3 and x=0). Therefore, the area of this trapezoid is:
(1/2)*(3+0)*(6+(-6)) = 0
2. Area of the region under the line segment BC within the interval 0 to 5:
The equation of line BC is y = (13/5)x - 6. To find the area under this line segment within the interval 0 to 5, we calculate the area of the corresponding triangle.
The base length is 5 units (the x-distance) and the height is 7 units (the y-value at x=5). Therefore, the area of this triangle is:
(1/2)*5*7 = 17.5
3. Area of the region under the line segment CD within the interval 5 to b:
The equation of line CD is y = (-14/3)x + 91/3. To find the area under this line segment within the interval 5 to b, we calculate the area of the corresponding trapezoid.
The base lengths are (b - 5) units (the x-distance) and the heights are (-14/3) units (the y-values at x=5 and x=b). Therefore, the area of this trapezoid is:
(1/2)*(b-5)*((-14/3) + (-14/3)) = -(7/3)*(b-5)
Now, we can find the integral of f(x) from -3 to b by summing up the areas of the three subintervals:
Integral of f(x) from -3 to b = Area under line segment AB + Area under line segment BC + Area under line segment CD
= 0 + 17.5 + -(7/3)*(b-5)
= 17.5 - (7/3)*(b-5)
So, the integral of f(x) from -3 to b is 17.5 - (7/3)*(b-5).