Evaluate the integral by interpreting it in terms of areas.
0
(3 +sqrt(1 − x2)) dx
−1
look at the graph of y = 3 + √(1-x^2) from -1 to 0
because of the symmetry we could go from 0 to 1.
so the area consists of a rectangle 3 units by 1 unit, with the curve of √(1-x^2) sitting on top
let's just find the area of that
= ∫ √(1-x^2) dx from 0 to 1
not sure if you know the procedure to integrate that but Wolfram can do it for us
http://www.wolframalpha.com/input/?i=integrate+%E2%88%9A(1-x%5E2)+dx+from+0+to+1
add the rectangle and you are done
geometrically the argument makes sense. But algebraically,
∫[a,b] f(x) dx = -∫[b,a] f(x) dx
To evaluate the integral ∫[−1,0] (3 + √(1 − x^2)) dx by interpreting it in terms of areas, we can consider the function (3 + √(1 − x^2)) as the height of a region and dx as the width.
First, let's sketch the function √(1 − x^2) to understand the region involved:
When x is between −1 and 1, the function √(1 − x^2) represents the upper half of a unit circle centered at the origin, with a radius of 1. So, the region is the area under the curve √(1 − x^2) for −1 ≤ x ≤ 0.
Next, let's split the integral into two parts:
∫[−1,0] (3 + √(1 − x^2)) dx = ∫[−1,0] 3 dx + ∫[−1,0] √(1 − x^2) dx
The first integral, ∫[−1,0] 3 dx, represents the area of a rectangle with a fixed height of 3 and a width of (0 − (−1)) = 1. Therefore, its value is 3 * 1 = 3.
The second integral, ∫[−1,0] √(1 − x^2) dx, represents the area under the curve √(1 − x^2) from x = −1 to x = 0. This region corresponds to the lower half of the unit circle. To find this area, we can use the formula for the area of a semi-circle, which is (π * r^2) / 2, where r is the radius.
In this case, the radius is 1, so the area of the semi-circle is (π * 1^2) / 2 = π / 2.
Therefore, the integral ∫[−1,0] (3 + √(1 − x^2)) dx can be evaluated as:
∫[−1,0] (3 + √(1 − x^2)) dx = ∫[−1,0] 3 dx + ∫[−1,0] √(1 − x^2) dx
= 3 + (π / 2)
= 3 + π / 2.
So the value of the integral is 3 + π / 2.
To evaluate the integral ∫[−1,0] (3 + sqrt(1 − x^2)) dx by interpreting it in terms of areas, we can consider the geometric interpretation of the integral.
The function f(x) = 3 + sqrt(1 − x^2) represents a semi-circle with a radius of 1 centered at (0,0), as the expression √(1 − x^2) denotes the upper half of the unit circle. The term 3 represents a horizontal line that is added to the semi-circle.
Interpreting the integral in terms of areas, we want to find the area between the curve and the x-axis over the interval [−1,0]. As the function is non-negative over the given interval, the integral represents the area bounded by the curve, the x-axis, and the vertical lines x = −1 and x = 0.
To evaluate this integral, we can split it into two parts: the area of the semi-circle and the area of the horizontal line.
1. Area of the Semi-Circle:
The semi-circle has a radius of 1, so its area can be calculated using the formula for the area of a circle, πr^2. In this case, the radius is 1, so the area of the semicircle is (1/2) * π * 1^2 = π/2.
2. Area of the Horizontal Line:
The horizontal line represented by the constant term 3 extends from x = −1 to x = 0. Since the line is parallel to the x-axis, its height remains constant over the interval. The length of the interval is 1, and the height of the line is 3, so the area of the rectangle formed by the line and the x-axis is 3 * 1 = 3.
Finally, we add the areas of the semi-circle and the rectangle:
Area = Area of Semi-Circle + Area of Rectangle = π/2 + 3.
Therefore, the value of the integral ∫[−1,0] (3 + sqrt(1 − x^2)) dx, interpreted as the area bounded by the curve, the x-axis, and the interval [−1,0] is π/2 + 3.