The graph of g consists of two straight lines and a semicircle as shown in the figure.There are two line segments and a semicircle connected on the x y-coordinate plane. The graph is labeled y = g(x). The first line segment starts at the point (0, 20), goes down and right, and ends on the x-axis at the point (10, 0). The semicircle starts at the point (10, 0), passes through (20, −10), and ends at the point (30, 0). The second line segment starts at the point (30, 0), goes up and right, and ends at the point (35, 5).Evaluate each integral by interpreting it in terms of areas.(a)10g(x) dx0100−50π (b)30g(x) dx1050π−50 (c)35g(x) dx0
Question
The graph of g consists of two straight lines and a semicircle as shown in the figure.There are two line segments and a semicircle connected on the x y-coordinate plane. The graph is labeled y = g(x). The first line segment starts at the point (0, 20), goes down and right, and ends on the x-axis at the point (10, 0). The semicircle starts at the point (10, 0), passes through (20, −10), and ends at the point (30, 0). The second line segment starts at the point (30, 0), goes up and right, and ends at the point (35, 5).Evaluate each integral by interpreting it in terms of areas.(a)10g(x) dx0100−50π (b)30g(x) dx1050π−50 (c)35g(x) dx0
Solution
To solve these integrals, we need to interpret them in terms of areas under the curve of the function g(x).
(a) ∫ from 0 to 10 g(x) dx
This integral represents the area under the curve from x=0 to x=10. Looking at the graph, this area is a triangle with base 10 and height 20. The area of a triangle is 1/2 * base * height, so the area is 1/2 * 10 * 20 = 100.
(b) ∫ from 10 to 30 g(x) dx
This integral represents the area under the curve from x=10 to x=30. This area is a semicircle with radius 10 (the distance from (20, -10) to (10, 0) or (30, 0)). The area of a semicircle is 1/2 * π * radius^2, so the area is 1/2 * π * 10^2 = 50π. However, because this area is below the x-axis, it is considered negative when calculating the integral, so the integral is -50π.
(c) ∫ from 0 to 35 g(x) dx
This integral represents the area under the curve from x=0 to x=35. This is the sum of the areas calculated in parts (a) and (b), plus the area of the triangle from x=30 to x=35. This last triangle has base 5 and height 5, so its area is 1/2 * 5 * 5 = 12.5. Adding up all these areas, the integral is 100 - 50π + 12.5.
Similar Questions
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