You are given the four points in the plane A = (-3,7), B = (-1,-6), C = (4,6), and D = (6,-1). The graph of the function f(x) consists of the three line segments AB, BC and CD. Find the integral interval -3 to 6 f(x)dx by interpreting the integral in terms of sums and/or differences of areas of elementary figures.
I got 18.4555, but it was wrong
i tried a different approach and got 1.5 but its still wrong
Did you graph the lines?
Let the points a,b,c,d be on the x-axis above or below A,B,C,D
AabB is a trapezoid with bases 7,6, height 2
BbcC is a rectangls 5x6
CcDd is a triangle of height 7 and width 2
However, part of the triangle is below the x-axis, making it a negative area.
Positive area is a triangle 6 by 12/7
Negative area is a trapezoid with bases 12/7 and 2, height 1
Add the areas:
[(7+6)/2 * 2] + [5*6] + [6 * 12/7 * 1/2] - [(12/7 + 2)/2*1]
= 13 + 30 + 36/7 - 13/7
= 43 - 23/7 = 39 5/7 = 39.71
better check my math here
no, the answer is 6, you separate the triangles, find the area and add them up
To find the integral of the function f(x) over the interval -3 to 6, we need to interpret it in terms of sums and/or differences of areas of elementary figures.
First, let's plot the given points A, B, C, and D on a coordinate plane to visualize the three line segments AB, BC, and CD.
A = (-3,7)
B = (-1,-6)
C = (4,6)
D = (6,-1)
Based on the given points, we can see that the graph of f(x) consists of three line segments: AB, BC, and CD.
To find the integral, we need to find the areas of the elementary figures formed by these line segments. We can divide the interval -3 to 6 into three sub-intervals: -3 to -1, -1 to 4, and 4 to 6.
The area of the elementary figure formed by the line segment AB can be expressed as the difference of the areas of two triangles: a larger triangle and a smaller triangle. The larger triangle has a base length of 2 units (-1 minus -3) and a height of 13 units (7 minus -6). Therefore, the area of the larger triangle is (1/2) * 2 * 13 = 13 square units. The smaller triangle has a base length of 2 units (-1 minus -3) and a height of 7 units (0 minus -6). Therefore, the area of the smaller triangle is (1/2) * 2 * 7 = 7 square units. The area of the elementary figure formed by AB is the difference between the areas of these triangles: 13 - 7 = 6 square units.
The area of the elementary figure formed by the line segment BC can also be expressed as the difference of the areas of two triangles. The larger triangle has a base length of 5 units (4 minus -1) and a height of 12 units (6 minus -6). Therefore, the area of the larger triangle is (1/2) * 5 * 12 = 30 square units. The smaller triangle has a base length of 3 units (4 minus 1) and a height of 6 units (0 minus -6). Therefore, the area of the smaller triangle is (1/2) * 3 * 6 = 9 square units. The area of the elementary figure formed by BC is the difference between the areas of these triangles: 30 - 9 = 21 square units.
The area of the elementary figure formed by the line segment CD can be expressed as the difference of the areas of two rectangles. The larger rectangle has a base length of 2 units (6 minus 4) and a height of 4 units (-1 minus -6). Therefore, the area of the larger rectangle is 2 * 4 = 8 square units. The smaller rectangle has a base length of 2 units (6 minus 4) and a height of 7 units (0 minus -7). Therefore, the area of the smaller rectangle is 2 * 7 = 14 square units. The area of the elementary figure formed by CD is the difference between the areas of these rectangles: 8 - 14 = -6 square units.
Now, let's sum up the areas of the elementary figures: 6 + 21 + (-6) = 21 square units.
Therefore, the integral of f(x) over the interval -3 to 6 is equal to 21 square units.
I apologize if my explanation was not clear enough. The result you obtained, 18.4555, is incorrect. The correct answer is 21 square units.