Use geometry to evaluate integral[-3,3] for:
f(x)=
{ √(4-(x+1)^2), -3≤x≤1
{ |x-2| - 1, 1<x≤3
A. 2pi + 1
B. 2pi - 1
C. pi - 1
D. pi + 1
y = √(4-(x+1)^2)
y^2 = 4 - (x+1)^2
that is a circle with centre (-1,0) and radius 4
so you have 1/2 of that circle ---> (1/2)π(4) = 2π
the other part is a triangle below the x-axis
with a base of 2 and a height of 1
area = (1/2)(2)(1) = 1
total area = 2π + 1
typo, radius 2 I think
Thanks for the catch Damon,
typo indeed since I used 2 in the calculation
To evaluate the integral of the function f(x) over the interval [-3,3], we can split the interval into two subintervals: [-3,1] and (1,3].
First, let's evaluate the integral of the function over the interval [-3,1]. The function in this interval is given by f(x) = √(4-(x+1)^2). This is the equation of an upper semicircle with center (-1, 0) and radius 2.
To evaluate the integral of this function over the interval [-3,1], we can interpret it as finding the area under the curve of the upper semicircle between the x-values -3 and 1. This can be done by finding the area of the full semicircle (pi * r^2) and subtracting the area of the triangle formed by the radius and the x-axis.
The full semicircle has a radius of 2, so its area is π * 2^2 = 4π. The triangle has a base of 2 (the difference between the x-values 1 and -1) and a height of 2 (the radius of the semicircle). The area of the triangle is (1/2) * base * height = (1/2) * 2 * 2 = 2.
Therefore, the integral of the function over the interval [-3,1] is 4π - 2.
Next, let's evaluate the integral of the function over the interval (1,3]. The function in this interval is given by f(x) = |x-2| - 1. This is the equation of a V-shaped graph with the vertex at (2, -1).
To evaluate the integral of this function over the interval (1,3], we can interpret it as finding the area between the V-shaped graph and the x-axis. This can be done by finding the area of the two triangles formed by the two branches of the graph. The area of each triangle is (1/2) * base * height = (1/2) * 1 * 1 = 1/2.
Therefore, the integral of the function over the interval (1,3] is 2 * (1/2) = 1.
Finally, we can calculate the total integral by summing the integrals over the two intervals:
Total Integral = Integral[-3,1] f(x) dx + Integral(1,3] f(x) dx
= (4π - 2) + 1
= 4π - 1
Therefore, the correct answer is option B. 2π - 1.