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.
Oh, dear! You're asking me to do some math? I'm more of a clown than a mathematician. But don't worry, I can still help.
To find the integral of f(x) from -3 to ?, we need to find the sum of the areas under the three line segments AB, BC, and CD.
Let's break it down step by step, shall we?
First, let's find the area under AB. The length of AB is 12 units (6 - (-6)). The height is 3 units (-3 - 0). So, the area under AB is 12 * 3 = 36 square units.
Next, let's find the area under BC. The length of BC is 13 units (7 - (-6)). The height is 5 units (5 - 0). So, the area under BC is 13 * 5 = 65 square units.
Finally, let's find the area under CD. The length of CD is 15 units (-7 - 8). The height is 13 units (7 - (-6)). So, the area under CD is 15 * 13 = 195 square units.
To find the integral of f(x) from -3 to ?, we need to sum up the areas under AB, BC, and CD. That gives us 36 + 65 + 195 = 296 square units.
And there you have it! The integral of f(x) from -3 to ? is 296 square units. Now, let's get back to making people laugh!
To find the integral of f(x) from -3 to [missing value], we can break down the function into three line segments: AB, BC, and CD. We'll find the area under each segment separately and then sum them up.
1. AB Line Segment:
The equation of the line passing through points A(-3,6) and B(0,-6) can be calculated using the slope-intercept form: y = mx + b.
First, find the slope (m):
m = (change in y) / (change in x) = (-6 - 6) / (0 - (-3)) = -12 / 3 = -4
Next, find the y-intercept (b):
Using the equation y = mx + b, plug in one point (A or B) and the slope (m) to solve for b.
6 = -4(-3) + b
6 = 12 + b
b = -6
The equation of the line AB is therefore: y = -4x - 6.
To find the area under AB, we need to compute the definite integral of the line segment from -3 to 0:
∫[from -3 to 0] (-4x - 6) dx
Integrating term by term:
∫[from -3 to 0] -4x dx - ∫[from -3 to 0] 6 dx
Solving the integrals:
[-2x^2] (from -3 to 0) - [6x] (from -3 to 0)
-2(0)^2 - (-2(-3)^2) - 6(0) - (-6(-3))
0 - (-18) - 0 + 18
0 + 18 + 0 + 18 = 36
The area under AB is 36.
2. BC Line Segment:
The equation of the line passing through points B(0,-6) and C(5,7) can be calculated in the same way.
Find the slope:
m = (change in y) / (change in x) = (7 - (-6)) / (5 - 0) = 13 / 5
Find the y-intercept (b):
Using the equation y = mx + b, plug in one point (B or C) and the slope (m) to solve for b.
-6 = (13/5)(0) + b
b = -6
The equation of the line BC is therefore: y = (13/5)x - 6.
To find the area under BC, we need to compute the definite integral of the line segment from 0 to 5:
∫[from 0 to 5] ((13/5)x - 6) dx
Integrating term by term:
∫[from 0 to 5] (13/5)x dx - ∫[from 0 to 5] 6 dx
Solving the integrals:
[(13/10)x^2] (from 0 to 5) - [6x] (from 0 to 5)
(13/10)(5^2) - (13/10)(0^2) - 6(5) + 6(0)
(13/10)(25) - 0 - 30 + 0
13(5/2) - 30 = 65/2 - 30 = -5/2
The area under BC is -5/2.
3. CD Line Segment:
The equation of the line passing through points C(5,7) and D(8,-7) can be calculated in the same way.
Find the slope:
m = (change in y) / (change in x) = (-7 - 7) / (8 - 5) = -14 / 3
Find the y-intercept (b):
Using the equation y = mx + b, plug in one point (C or D) and the slope (m) to solve for b.
7 = (-14/3)(5) + b
b = 63/3 = 21
The equation of the line CD is therefore: y = (-14/3)x + 21.
To find the area under CD, we need to compute the definite integral of the line segment from 5 to 8:
∫[from 5 to 8] ((-14/3)x + 21) dx
Integrating term by term:
∫[from 5 to 8] (-14/3)x dx + ∫[from 5 to 8] 21 dx
Solving the integrals:
[(-7/3)x^2] (from 5 to 8) + [21x] (from 5 to 8)
(-7/3)(8^2) - (-7/3)(5^2) + 21(8) - 21(5)
(-7/3)(64) - (-7/3)(25) + 168 - 105
-448/3 + 175/3 + 168 - 105 = -273/3 + 273 - 105 = -273/3 + 168 = -105/3
The area under CD is -105/3.
Finally, to find the integral of f(x) from -3 to [missing value], we sum up the areas under each line segment:
∫[from -3 to [missing value]] f(x) dx = Area under AB + Area under BC + Area under CD
= 36 + (-5/2) + (-105/3) = 72/2 - 15/2 - 105/3
= 36 - 15/2 - 35 = 72/2 - 15/2 - 70/2
= 36/2 - 15/2 - 70/2 = 18 - 15 - 35 = -32.5
Therefore, the integral of f(x) from -3 to [missing value] is -32.5.