Use geometry to evaluate the integral from negative 4 to 4 of the square root of the quantity 16 minus x squared, dx.
pi divided by 4
4π
8π
16π
The curve is a semicircle of radius 4
Looks like 8π to me.
Aha, we are diving into the world of integrals and geometry! To evaluate this integral, let's first do a little bit of geometry.
The intrepid expression inside the square root, 16 - x^2, is actually that of a circle centered at the origin with a radius of 4. The integral from -4 to 4 is asking us to find the area under the curve within this range. Sounds interesting, doesn't it?
Now, when you integrate the square root of 16 - x^2, it turns out to be the formula for the upper half of this circle. In other words, we're calculating half of the area of this circle!
And what is the formula for the area of a full circle? Ah, that's π * r^2, where r is the radius. So, since we're calculating half of the circle, the area is (1/2) * π * 4^2, which simplifies to 8π.
So, my friend, the correct answer is 8π. Don't worry, I've got pi jokes to spare if you need 'em!
To evaluate the integral, we can use geometry and apply properties of circles.
First, let's consider the function inside the square root: 16 - x^2. This represents the equation of a circle centered at the origin with a radius of 4. We can rewrite this equation as: x^2 + y^2 = 16.
Next, let's graph the region defined by the integral limits from -4 to 4. This represents the interval on the x-axis from -4 to 4. Using geometry, we can visualize that this interval corresponds to a length of 8 units on the x-axis.
Now, let's evaluate the integral using geometry. We want to find the area under the graph of the square root of (16 - x^2) over the interval from -4 to 4. This area can be represented by a quarter of the circle centered at the origin with a radius of 4.
The area of a circle is given by the formula A = πr^2. In this case, the quarter of the circle corresponds to 1/4 of the total area. So, the integral we have is equal to (1/4) * π * 4^2.
Simplifying this expression, we have (1/4) * π * 16, which simplifies to 4π.
Therefore, the value of the integral from -4 to 4 of the square root of (16 - x^2), dx, is equal to 4π.
To evaluate the integral, let's start by visualizing the region defined by the equation y = √(16 - x²).
The integrand represents the upper half of a circle with radius 4 centered at the origin. We need to find the area of this region between x = -4 and x = 4.
We can break down the region into two symmetrical halves, from x = -4 to x = 0 and from x = 0 to x = 4. The area under the curve in each half is equal.
So, we just need to calculate the area of one-half and then double it. Let's evaluate the integral from -4 to 0 first:
∫[from -4 to 0] √(16 - x²) dx
This integral represents the area under the curve in the first quadrant. We can compute it by recognizing that it is one-fourth of the area of a full circle with radius 4.
The formula for the area of a circle is A = πr², where r is the radius. Therefore, the area of a circle with radius 4 is A = π(4)² = 16π.
Since we are looking for one-fourth of this area, the area of the first quadrant is 16π / 4 = 4π.
Now let's calculate the area from x = 0 to x = 4:
∫[from 0 to 4] √(16 - x²) dx
This represents the area under the curve in the second quadrant. Since it is symmetrical to the first quadrant, the area in the second quadrant is also 4π.
Finally, we can find the total area by summing the areas of both quadrants:
Total Area = 4π (first quadrant) + 4π (second quadrant) = 8π.
Therefore, the integral from -4 to 4 of √(16 - x²) dx is equal to 8π.
So, the correct answer is 8π.