Problem 72 - Calculus Early Transcendentals

Question

On the interval [0,2] , the graphs of \(f(x)=x^{2} / 3\) and \(g(x)=x^{2}\left(9-x^{2}\right)^{-1 / 2}\) have similar shapes. a. Find the area of the region bounded by the graph of \(f\) and the \(x\) -axis on the interval [0,2] b. Find the area of the region bounded by the graph of \(g\) and the \(x\) -axis on the interval [0,2] c. Which region has greater area?

Step-by-Step Solution

Answer

Answer: The region bounded by g(x) and the x-axis has a greater area.

Step 1: Area of the region bounded by f(x) and the x-axis

To find the area of the region bounded by f(x)=x^2/3 and the x-axis on the interval [0,2], we need to find the definite integral of f(x) on the interval [0,2]. We use the integral of the function to find the area: $$\int_{0}^{2} \frac{x^2}{3}\,dx$$

Step 2: Solve the definite integral of f(x)

We integrate the given function with respect to x: $$\int \frac{x^2}{3}\,dx = \frac{1}{3} \int x^2\,dx = \frac{1}{3} \cdot \frac{x^3}{3} + C = \frac{x^3}{9} + C$$ Now we evaluate the definite integral: $$\int_{0}^{2} \frac{x^2}{3}\,dx = \left[\frac{x^3}{9}\right]_0^2 = \frac{2^3}{9} - \frac{0^3}{9} = \frac{8}{9}$$

Step 3: Area of the region bounded by g(x) and the x-axis

To find the area of the region bounded by g(x)=x^2(9-x^2)^{-1/2} and the x-axis on the interval [0,2], we need to find the definite integral of g(x) on the interval [0,2]. We use the integral of the function to find the area: $$\int_{0}^{2} x^2\left(9-x^{2}\right)^{-1 / 2}\,dx$$

Step 4: Solve the definite integral of g(x)

We integrate the given function with respect to x, which is not as straightforward as finding the definite integral of f(x). One technique to tackle this integration is by using substitution. Let $$u = 9-x^2$$ Differentiating with respect to x, we get: $$\frac{du}{dx} = -2x$$ Rearranging to find dx: $$dx = -\frac{du}{2x}$$ Substitute the variables u and dx in the integral: $$\int -x^2(9-x^2)^{-1/2} \cdot \frac{du}{2x}$$ Now we can simplify the integral to: $$-\frac{1}{2} \int u^{-1/2}\,du$$ Next, integrate with respect to u: $$-\frac{1}{2} \int u^{-1/2}\,du = -\frac{1}{2} \cdot 2u^{1/2} + C = -u^{1/2} + C$$ Now substitute back x: $$- (9-x^2)^{1/2} + C$$ Evaluate the definite integral: $$\int_{0}^{2} x^2\left(9-x^{2}\right)^{-1 / 2}\,dx = \left[ -(9-x^2)^{1/2}\right]_0^2 = - (9-2^2)^{1/2} + (9-0^2)^{1/2}= -1 + 3 = 2$$

Step 5: Compare the areas

Area of the region bounded by f(x) and the x-axis is \(\frac{8}{9}\), and area of the region bounded by g(x) and the x-axis is \(2\). Compare the two areas: $$\frac{8}{9} < 2$$ Therefore, the region bounded by g(x) and the x-axis has a greater area.