You are given the four points in the plane A=(−1,−8)A=(−1,−8), B=(4,5)B=(4,5), C=(8,−2)C=(8,−2), and D=(12,2)D=(12,2). The graph of the function f(x)f(x) consists of the three line segments AB, BC and CD. Find the integral ∫12−1f(x)dx by interpreting the integral in terms of sums and/or differences of areas of elementary figures.
Well, well, well, looks like we've got ourselves some line segments to play with! And now we want to find the integral of this curvy function f(x)? I guess it's time to break out my clown calculator!
First, let's analyze this problem step by step. We have three line segments, AB, BC, and CD, and we want to find the integral of f(x) over the interval [-1,12]. So, we need to find the area under the curve f(x) between -1 and 12.
To do this, we can divide the interval [-1,12] into smaller intervals and approximate the area under each segment using basic shapes like rectangles or triangles. Let's go segment by segment:
Segment AB: We can approximate this segment with a triangle. The base of the triangle is 4 units and the height is 13 units (the difference in y-coordinates of A and B). So, the area of this triangle is (1/2) * 4 * 13 = 26 square units.
Segment BC:Similarly, we can approximate this segment with a triangle. The base of the triangle is 4 units and the height is 7 units (the difference in y-coordinates of B and C). So, the area of this triangle is (1/2) * 4 * 7 = 14 square units.
Segment CD: Again, we can approximate this segment with a triangle. The base of the triangle is 4 units and the height is 4 units (the difference in y-coordinates of C and D). So, the area of this triangle is (1/2) * 4 * 4 = 8 square units.
Now, let's add up the areas of these approximations:
Total area ≈ 26 + 14 + 8 = 48 square units.
Since we want to find the integral of f(x) over the interval [-1,12], the value of the integral ∫12−1f(x)dx will be approximately equal to 48.
So, there you have it, my friend! The answer to the integral ∫12−1f(x)dx is approximately 48 (square units). Of course, this is just an approximation using basic shapes, but it should give you a decent estimate. Hope that brings a smile to your face!
To find the integral ∫12−1f(x)dx by interpreting it in terms of sums and differences of areas of elementary figures, we need to break down the function into the three line segments AB, BC, and CD.
First, let's calculate the equation of each line segment:
Line segment AB:
The slope of AB can be found using the formula:
m = (y2 - y1) / (x2 - x1)
Using the points A and B:
m = (5 - (-8)) / (4 - (-1))
m = 13 / 5
The equation of AB can be represented as:
y = mx + b
where m is the slope and b is the y-intercept.
We need to find the y-intercept (b) by substituting the coordinates of point A into the equation:
-8 = (13/5)(-1) + b
-8 = -13/5 + b
-8 + 13/5 = b
(-40/5) + (13/5) = b
-27/5 = b
Therefore, the equation of line segment AB is:
y = (13/5)x - 27/5
Line segment BC:
Using points B and C:
m = (-2 - 5) / (8 - 4)
m = -7 / 4
The equation of BC:
y = (-7/4)x + b
Substituting point B into the equation:
5 = (-7/4)(4) + b
5 = -7 + b
5 + 7 = b
12 = b
Thus, the equation of line segment BC is:
y = (-7/4)x + 12
Line segment CD:
Using points C and D:
m = (2 - (-2)) / (12 - 8)
m = 4 / 4
m = 1
The equation of CD:
y = 1x + b
Substituting point C into the equation:
-2 = 8 + b
b = -10
The equation of line segment CD becomes:
y = x - 10
Now, to find the integral ∫12−1f(x)dx, we need to find the areas of the elementary figures formed by these line segments.
- Consider the region formed by line segment AB and the x-axis from x = -1 to x = 4. This region is a trapezoid with height h1 = 13/5 - (-8) = 13/5 + 40/5 = 53/5 and bases b1 = 4 - (-1) = 5. The area of this trapezoid is given by:
A1 = (1/2) * (b1 + b2) * h = (1/2) * (5 + 53/5) * 53/5 = 212/5
- Consider the triangle formed by line segment BC and the x-axis from x = 4 to x = 8. The height of this triangle is given by h2 = 12 - (-10) = 22, and the base is b2 = 8 - 4 = 4. The area of this triangle is given by:
A2 = (1/2) * b2 * h2 = (1/2) * 4 * 22 = 44
- Finally, consider the region formed by line segment CD and the x-axis from x = 8 to x = 12. This region is a trapezoid with height h3 = 12 - (-2) = 14 and bases b3 = 12 - 8 = 4. The area of this trapezoid is given by:
A3 = (1/2) * (b3 + b4) * h3 = (1/2) * (4 + 4) * 14 = 56
To find the integral ∫12−1f(x)dx, we sum up the areas of these elementary figures:
∫12−1f(x)dx = A1 + A2 + A3 = 212/5 + 44 + 56 = 64/5
Therefore, the value of the integral ∫12−1f(x)dx, by interpreting it in terms of sums and differences of areas of elementary figures, is 64/5.