Evaluate the integral by interpreting it in terms of areas. In other words, draw a picture of the region the integral represents, and find the area using high school geometry.
interval 0 to 11 |3x-15|dx
From x = 0 to 5, |3x-15| goes from 15 to 0, as a straight line.
From x = 5 to 11, |3x-15| goes from 0 to 18, as a straight line.
Compute the areas of the two triangular regions separately, and add them.
(1/2)*5*15 + (1/2)*6*18 = ______
To evaluate the given integral ∫(0 to 11) |3x - 15| dx, we will interpret it in terms of areas and use high school geometry to find the area.
Step 1: Understanding the Integral
The expression |3x - 15| represents the absolute value of the function 3x - 15. By evaluating this integral, we are essentially finding the area between the curve represented by the absolute value function and the x-axis, within the interval from 0 to 11.
Step 2: Drawing the Graph
To visualize the region, we will first graph the function |3x - 15|. This function has two linear segments, one with a positive slope and another with a negative slope. Let's break it down into two separate segments and determine the critical point where the function changes slope.
For 3x - 15 = 0:
3x = 15
x = 5
Segment 1: For x ≤ 5, the function becomes: 3x - 15.
The graph of this segment is a line passing through the point (0, -15) with a slope of 3.
Segment 2: For x ≥ 5, the function becomes: -(3x - 15) = -3x + 15.
The graph of this segment is a line passing through the point (5, 0) with a slope of -3.
Step 3: Visualizing the Regions
Now we will divide the interval [0, 11] into three parts: [0, 5], [5, 8], and [8, 11].
Region 1: [0, 5]
In this region, the function |3x - 15| is equivalent to 3x - 15.
The area of this region is the area under the line with a positive slope.
Since the line is a triangle, its area can be calculated using the formula for the area of a triangle: A = (base * height) / 2.
Here, the base is (5 - 0) = 5, and the height is the y-coordinate at x = 5, which is (3 * 5 - 15) = 0.
Therefore, the area of Region 1 would be (5 * 0) / 2 = 0.
Region 2: [5, 8]
In this region, the function |3x - 15| is equivalent to -(3x - 15) = -3x + 15.
The area of this region is the area above the line with a negative slope.
Again, this is a triangle, so we can calculate its area using the formula for the area of a triangle.
The base is (8 - 5) = 3, and the height is the y-coordinate at x = 5, which is (-(3 * 5 - 15)) = 0.
Thus, the area of Region 2 is (3 * 0) / 2 = 0.
Region 3: [8, 11]
In this region, the function |3x - 15| is equivalent to 3x - 15.
The area of this region is the area below the line with a positive slope.
Similar to Region 1, this is a triangle, and we can use the formula for the area of a triangle.
The base is (11 - 8) = 3, and the height is the y-coordinate at x = 5, which is (3 * 5 - 15) = 0.
So, the area of Region 3 becomes (3 * 0) / 2 = 0.
Step 4: Evaluating the Integral
Now that we have determined the areas of the three regions, we sum them up to find the total area.
The total area is calculated as the sum of the areas of all the regions:
Area = Area of Region 1 + Area of Region 2 + Area of Region 3
= 0 + 0 + 0
= 0
Therefore, the integral of ∫(0 to 11) |3x - 15| dx is equal to 0, as the total area enclosed by the curve and the x-axis within the interval [0, 11] is zero.
To evaluate the integral ∫[0, 11] |3x-15| dx in terms of areas, we can interpret it as finding the area between the curve |3x-15| and the x-axis over the interval [0, 11].
First, let's break down the integral into two separate integrals, one over the interval [0, 5] and another over the interval [5, 11], since the absolute value function |3x-15| changes direction at x = 5:
∫[0, 11] |3x-15| dx = ∫[0, 5] |3x-15| dx + ∫[5, 11] |3x-15| dx
Now, let's focus on the interval [0, 5]. The function |3x-15| can be simplified as:
|3x-15| = 3x-15, for 0 ≤ x ≤ 5
To find the area under this part of the curve, we can integrate the function 3x-15 over the interval [0, 5]:
∫[0, 5] (3x-15) dx
To evaluate this integral, we can use the power rule of integration:
∫[0, 5] (3x-15) dx = [(3/2)x^2 - 15x] evaluated from 0 to 5
= [(3/2)(5)^2 - 15(5)] - [(3/2)(0)^2 - 15(0)]
= [(3/2)(25) - 75] - [0 - 0]
= [75/2 - 75] - [0]
= (75/2 - 75)
= -75/2
Therefore, the area under the curve |3x-15| for 0 ≤ x ≤ 5 is -75/2.
Next, let's consider the interval [5, 11]. The function |3x-15| can be simplified as:
|3x-15| = -(3x-15), for 5 ≤ x ≤ 11
To find the area under this part of the curve, we can integrate the function -(3x-15) over the interval [5, 11]:
∫[5, 11] -(3x-15) dx
Using the power rule of integration, we can simplify this integral:
∫[5, 11] -(3x-15) dx = [-((3/2)x^2 - 15x)] evaluated from 5 to 11
= [-(3/2)(11)^2 + 15(11)] - [-(3/2)(5)^2 + 15(5)]
= [-((3/2)(121) - 165)] - [-(3/2)(25) + 75]
= [-363/2 + 165] - [-75/2 + 75]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
= [165 - 363/2] - [75 - 75/2]
After evaluating the above expression, we find the area under the curve |3x-15| for 5 ≤ x ≤ 11.