Evaluate the integral by interpreting it in terms of areas.
bounds from -1 to 0.
3+sqrt(1−x2) dx
y=sqrt(1-x^2) is the equation for the top half of a circle of radius 1.
So, this would be a rectangle + 1/4 circle
To evaluate the integral ∫(3+sqrt(1−x^2)) dx over the interval [-1, 0], we can interpret this integral in terms of areas.
First, let's understand the integrand √(1−x^2). This is the equation of a semicircle centered at (0,0) with a radius of 1. The 3 in the integrand, 3+√(1−x^2), represents a constant shift vertically upwards by 3 units.
To evaluate the integral, we need to find the area enclosed between the curve and the x-axis in the given interval. Since the integrand is always positive, the area will be the same as the definite integral.
We can break down the interval [-1, 0] into two separate regions: the negative segment and the positive segment.
1. Negative segment: x ranges from -1 to 0.
In this range, the integrand 3+√(1−x^2) is always positive, resembling the upper half of the semicircle. The area under the curve can be calculated as the area of the semicircle with radius 1, plus the constant shift of 3.
The area of the semicircle is (π * r^2) / 2 = (π * 1^2) / 2 = π/2.
Adding the constant shift of 3, the total area under the curve in the negative segment is π/2 + 3.
2. Positive segment: x ranges from 0 to -1.
In this range, the integrand 3+√(1−x^2) is negative, resembling the lower half of the semicircle. However, since we are integrating over a negative segment, we need to take the absolute value of the integrand to find the positive area.
The absolute value of 3+√(1−x^2) is simply 3-√(1−x^2), which represents the lower half of the semicircle.
Similarly, the area under this curve can be calculated as the area of the semicircle with radius 1, minus the constant shift of 3.
The area of the semicircle is (π * r^2) / 2 = (π * 1^2) / 2 = π/2.
Subtracting the constant shift of 3, the total area under the curve in the positive segment is π/2 - 3.
Now, to find the total area enclosed between the curve and the x-axis, we sum up the areas of the two segments:
Total area = (Area in the negative segment) + (Area in the positive segment)
= (π/2 + 3) + (π/2 - 3)
= π
Hence, the value of the integral ∫(3+sqrt(1−x^2)) dx over the interval [-1, 0] is equal to π.