Use geometry to evaluate the integral from negative 3 to 3 of f of x, dx for f of x equals the square root of the quantity 4 minus the square of the quantity x plus 1 for x is between negative 3 and 1 including negative 3 and 1, and equals the absolute value of the quantity (x minus 2) minus 1 for x is greater than 1 and less than or equal to 3.
2pi-1
If I read all those words correctly, you mean
∫[-3,3] f(x) dx
where
f(x) =
f1: √(4-√(x+1)) for -3 <= x <= 1
f2: |x-2|-1 for 1<x<3
Unfortunately, for x < -1 the first piece is complex, not real.
Anyway, fix the presumed typo and just plug in the pieces
∫[-3,1] f1(x) dx + ∫[1,3] f2(x) dx
To evaluate the integral using geometry, we can break it down into two separate regions based on the given piecewise definition of f(x).
1. Region 1: When x is between -3 and 1 (including -3 and 1)
For this region, f(x) is defined as the square root of the quantity 4 - (x+1)^2. The graph of this function represents a semi-circle since it involves the square root of a quadratic expression.
To evaluate the integral over this region, we can calculate the area of the semi-circle.
The radius of the semi-circle is determined by the expression inside the square root: 4 - (x+1)^2. To find the radius at the boundaries of this region, we substitute the values of x = -3 and x = 1 into the expression:
Radius at x = -3: √(4-(-3+1)^2) = √(4-(-2)^2) = √(4-4) = 0
Radius at x = 1: √(4-(1+1)^2) = √(4-2^2) = √(4-4) = 0
Since the radius is zero at both boundaries, the shape of the curve for this region is actually a horizontal line at y = 0.
Therefore, the area under this horizontal line is equal to 0, and the integral over this region is 0.
2. Region 2: When x is greater than 1 and less than or equal to 3
For this region, f(x) is defined as the absolute value of (x-2) - 1. The graph of this function represents a V-shape with the vertex at the point (2, -1).
To evaluate the integral over this region, we can calculate the area of the V-shape.
Since the given function expression involves an absolute value, two triangles will be formed.
The base of each triangle is the difference between the x-values: 3 - 1 = 2.
The height of each triangle is determined by substituting the x-values into the function expression:
Height at x = 1: |1 - 2 - 1| = |-2| = 2
Height at x = 3: |3 - 2 - 1| = |0| = 0
The area of each triangle can be found using the formula for the area of a triangle: (base * height) / 2.
Area of triangle at x = 1: (2 * 2) / 2 = 2
Area of triangle at x = 3: (2 * 0) / 2 = 0
To calculate the integral over this region, we need to add the areas of both triangles:
Integral over this region = Area of triangle at x = 1 + Area of triangle at x = 3 = 2 + 0 = 2
Therefore, the integral from -3 to 3 of f(x) can be evaluated as 0 + 2 = 2.
To evaluate the given integral using geometry, we can split the interval [-3, 3] into two parts based on the definition of the function \(f(x)\). The first part is from -3 to 1, and the second part is from 1 to 3.
Let's begin with the first part, from -3 to 1.
Step 1: Find the area of the region under the curve of \(f(x)\), which is the square root of (4 - (x + 1)^2), for x in the range -3 to 1. Since the function is always non-negative in this interval, we can interpret this area as the area between the curve and the x-axis.
To find the area of this region, we can use basic geometry. The curve is a semi-circle centered at (-1,0), with a radius of 2. The region under the curve can be divided into a semi-circle and a right triangle.
The area of the semi-circle is given by half the product of the radius squared and \(\pi\). So, the area of the semi-circle in this case is \((1/2)\cdot 2^2 \cdot \pi = 2\pi\).
The area of the right triangle is given by half the product of the base and height. The base of the triangle is 2 (the difference between 1 and -1), and the height is the y-coordinate at x = 1, which is \(\sqrt{4 - (1 + 1)^2} = \sqrt{4 - 4} = 0\). Therefore, the area of the right triangle is \((1/2)\cdot 2 \cdot 0 = 0\).
Adding the areas of the semi-circle and the triangle, we get a total area of 2\(\pi\).
Step 2: Determine the sign of the function for this interval (-3 to 1). Since the given function is non-negative for this region, the area we calculated (2\(\pi\)) represents the integral of \(f(x)\) from -3 to 1.
Now, let's move to the second part, from 1 to 3.
Step 1: Find the area of the region under the curve of \(f(x)\), which is the absolute value of ((x - 2) - 1), for x in the range 1 to 3. Since the function is non-negative for this interval, we can interpret this area as the area between the curve and the x-axis.
To find the area of this region, we can again use basic geometry. The curve is a line segment connecting the points (1, 0) and (3, 2).
The area of a right triangle is half the product of the base and the height. The base of the triangle is 2 (the difference between 3 and 1), and the height is the y-coordinate at x = 3, which is the absolute value of ((3 - 2) - 1) = 0. Therefore, the area of the right triangle is \((1/2)\cdot 2 \cdot 0 = 0\).
Since the function is non-negative for this interval, the area of the right triangle (0) represents the integral of \(f(x)\) from 1 to 3.
Step 2: Determine the sign of the function for this interval (1 to 3). Since the function is non-negative for this region, the area we calculated (0) represents the integral of \(f(x)\) from 1 to 3.
Finally, we can find the total integral of \(f(x)\) from -3 to 3 by summing up the individual integrals for the two intervals:
\(\int_{-3}^{3} f(x) dx = \int_{-3}^{1} f(x) dx + \int_{1}^{3} f(x) dx\)
\(\int_{-3}^{3} f(x) dx = 2\pi + 0\)
\(\int_{-3}^{3} f(x) dx = 2\pi\)
Therefore, the integral of \(f(x)\) from -3 to 3 is equal to \(2\pi\).