Sketch the region given by the definite integral. Use geometric shapes and formulas to evaluate the integral (a > 0, r > 0).

r
∫ sqrt(r^2 - x^2) dx
-r

While I recognize that this looks similar to a circle function, I'm not sure how to graph and evaluate this definite integral because the lower/upper bounds is throwing me off because it's not giving us a number, but a variable.

How could we approach this problem? I know there's tons of calculators that can calculate the integral by finding the anti derivative, but this is not the case for this problem specifically.

Any help is greatly appreciated!

Should we assume r = 1?

Do not assume r=1. But, it is a constant. And you are right, the graph is a sem-circle, so the area is π/2 r^2

Not sure where the a comes in, but to do the integral, now is the time to start getting used to trig substitutions. Let

x = r sinθ
r^2-x^2 = r^2(1-sin^2θ) = r^2 cos^2θ
dx = r cosθ dθ

Now you have

∫[-π/2,π/2] r cosθ (r cosθ dθ)
= r^2 ∫[-π/2,π/2] cos^2θ dθ
= r^2/2 ∫[-π/2,π/2] (1+cos2θ) dθ
= r^2/2 (θ + 1/2 sin2θ) [[-π/2,π/2]
= r^2/2 [(π/2+0)-(-π/2+0)]
= πr^2/2

To sketch the region given by the definite integral and evaluate it using geometric shapes and formulas, we need to understand the equation and the bounds of the integral.

The equation in the integral represents a semicircle of radius 'r' centered at the origin (0, 0).

Let's start by sketching the region. Since the integral bounds are defined in terms of 'r', we can denote the lower bound as '-r' and the upper bound as 'r'. This indicates that we are considering the portion of the semicircle that lies between the points (-r, 0) and (r, 0).

Next, let's evaluate the integral step-by-step.

∫ sqrt(r^2 - x^2) dx

We can first simplify the equation by recognizing that it represents the equation of the upper half of a circle. Applying the formula for the area of a semicircle, we have:

Area = (1/2)πr^2

Now, let's calculate the area of the region using the definite integral:

Area = ∫[from -r to r] sqrt(r^2 - x^2) dx

To evaluate this integral, we can substitute x = rsin(t):

dx = rcos(t) dt

When x = -r, t = -π/2, and when x = r, t = π/2.

The new integral becomes:

Area = ∫[from -π/2 to π/2] sqrt(r^2 - (rsin(t))^2) (rcos(t)) dt

Simplifying this further:

Area = ∫[from -π/2 to π/2] sqrt(r^2 - r^2sin^2(t)) (rcos(t)) dt

Area = ∫[from -π/2 to π/2] sqrt(r^2(1 - sin^2(t))) (rcos(t)) dt

Applying the identity sin^2(t) + cos^2(t) = 1, we can simplify to:

Area = ∫[from -π/2 to π/2] sqrt(r^2 cos^2(t)) (rcos(t)) dt

Area = ∫[from -π/2 to π/2] r^2 cos^2(t) dt

Using the double-angle identity for cosine, we have:

Area = ∫[from -π/2 to π/2] r^2 (1 + cos(2t))/2 dt

Now we can integrate each term of the integral separately:

Area = (r^2/2) ∫[from -π/2 to π/2] (1 + cos(2t)) dt

Integrating, we get:

Area = (r^2/2) [t + (sin(2t))/2] | from -π/2 to π/2

Evaluating at the upper and lower bounds, we obtain:

Area = (r^2/2) [(π/2 + sin(π)) - (-π/2 + sin(-π))]

We know that sin(π) = sin(-π) = 0, so this simplifies to:

Area = (r^2/2) [(π/2) - (-π/2)]

Area = (r^2/2) π

Therefore, the region given by the definite integral is an area of (r^2/2) π.

To better understand the region given by the definite integral and evaluate it using geometric shapes and formulas, we can visualize and relate it to a familiar geometric figure called a semi-circle.

The definite integral you provided, ∫(r to -r) sqrt(r^2 - x^2) dx, represents the area under the curve of the function y = sqrt(r^2 - x^2) from x = -r to x = r.

To graphically sketch and evaluate the region, we can start by visualizing a Cartesian coordinate plane. Draw the x-axis horizontally and the y-axis vertically.

Now, consider the equation of a circle with radius r centered at the origin:
x^2 + y^2 = r^2.

By isolating y in the equation above, we get:
y = sqrt(r^2 - x^2).

This equation corresponds to the upper half of the circle. The definite integral you provided calculates the area under this upper half of the circle between x = -r and x = r.

To visualize this, draw the semi-circle with the appropriate radius, r, on the coordinate plane, such that it touches the x-axis at (r, 0) and (-r, 0). Shade in the area between the x-axis and the semi-circle.

By geometry, we know that the area of a semi-circle is given by:
A_s = π * r^2 / 2.

So, to evaluate the definite integral, we can calculate the area of the shaded region, which corresponds to half of the area of the full circle of radius r.

Hence,
∫(r to -r) sqrt(r^2 - x^2) dx = half the area of the circle of radius r = π * r^2 / 2.

Therefore, the evaluation of the definite integral is π * r^2 / 2.

By understanding the geometric interpretation of the definite integral and relating it to the area of a semi-circle, we can sketch the region and evaluate the integral without relying on calculators or finding the antiderivative of the function.