Problem 25 - Calculus Single Variable

Question

Find the area of the region that is bounded by the graphs of \(y=f(x)\) and \(y=g(x)\) for \(x\) between the abscissas of the two points of intersection. $$ f(x)=x^{2}+5 \quad g(x)=2 x^{2}+1 $$

Step-by-Step Solution

Answer

The area of the region is \( \frac{32}{3} \) square units.

Step 1: Identify Points of Intersection

To find the region bounded by the two curves, we first need their points of intersection. This involves setting the two equations equal to each other: \ \( x^2 + 5 = 2x^2 + 1 \) \ Solve the equation for \( x \) by rearranging it: \ \( x^2 - 2x^2 + 5 - 1 = 0 \) \ Which simplifies to: \ \( -x^2 + 4 = 0 \) \ Rearranging gives: \ \( x^2 = 4 \), so \( x = \ \pm 2 \). The points of intersection are \( x = -2 \) and \( x = 2 \).

Step 2: Set Up the Integral for Area Calculation

Next, we calculate the area between the curves from \( x = -2 \) to \( x = 2 \). The area \( A \) is given by the integral: \ \[ A = \int_{-2}^{2} [(f(x) - g(x))] \, dx \] \ Here, \( f(x) = x^2 + 5 \) and \( g(x) = 2x^2 + 1 \). Substitute these into the integral: \ \[ A = \int_{-2}^{2} [(x^2 + 5) - (2x^2 + 1)] \, dx \] \ Simplify the expression inside the integral: \ \[ A = \int_{-2}^{2} [-x^2 + 4] \, dx \].

Step 3: Evaluate the Integral

We evaluate the integral \( \int_{-2}^{2} (-x^2 + 4) \, dx \). Integrate the function term-by-term: \ \[ \int ( -x^2 + 4) \, dx = \int -x^2 \, dx + \int 4 \, dx \] \ The antiderivatives are: \ \[ = -\frac{x^3}{3} + 4x \]. \ Evaluating from \(-2\) to \(2\): \ \[ \left[ -\frac{x^3}{3} + 4x \right]_{-2}^{2} \].

Step 4: Compute Definite Integral

Substitute the bounds into the antiderivative: \ \[ = \left( -\frac{2^3}{3} + 4 \times 2 \right) - \left( -\frac{(-2)^3}{3} + 4 \times (-2) \right) \] \ Simplify: \ \[ = \left( -\frac{8}{3} + 8 \right) - \left( \frac{8}{3} - 8 \right) \]. \ Calculate each: \ \[ = \left( -\frac{8}{3} + \frac{24}{3} \right) - \left( \frac{8}{3} - \frac{24}{3} \right) \] \ \[ = \left( \frac{16}{3} \right) - \left( -\frac{16}{3} \right) \]. \ \[ = \frac{16}{3} + \frac{16}{3} = \frac{32}{3} \].

Step 5: Finalize the Area Result

The total bounded area calculated is \( \frac{32}{3} \). Hence, the area of the region bounded by the graphs of \( y = f(x) \) and \( y = g(x) \) is \( \frac{32}{3} \) square units.

Key Concepts

Definite Integral

The concept of a definite integral is crucial in calculus for finding areas under curves. When integrating a function over a specified interval, we are essentially adding up an infinite number of infinitesimally small areas to find the total area between the curve and the x-axis over that interval. In our exercise, we used a definite integral to find the area between two curves over an interval from -2 to 2.

• We started by setting up the integral \( \int_{-2}^{2} [(f(x) - g(x))] \, dx \)
• Where \(f(x)\) and \(g(x)\) are the equations of the curves \( y = x^2 + 5 \) and \( y = 2x^2 + 1 \) respectively.

This setup directly calculates the area between the two curves by integrating the difference \( f(x) - g(x) \) over the specified interval. The definite integral helps us determine the exact size of this area, giving us the measure in terms of square units.

Area Between Curves

Finding the area between curves is a common application of definite integrals in calculus. It involves computing the area between two function graphs, over a given interval. Here, the area between the curves of \(y = x^2 + 5\) and \(y = 2x^2 + 1\) was the region of interest.

• The area was calculated by integrating the vertical distance \(f(x) - g(x)\) over the intersection interval determined earlier, from \(-2\) to \(2\).
• First, we simplified the expression \(-x^2 + 4\) and then integrated it

Upon simplifying, it is crucial to ensure we perform correct substitutions and follow steps meticulously to solve for the definite integral. Understanding this process allows us to accurately represent and compute the physical space contained between the graphs on the Cartesian plane.

Points of Intersection

Before calculating the area between curves, it is essential to find the intersection points. These points tell us over what range of x-values our two functions cross and outline the interval of integration. The intersection of two equations, like in our problem, involves:

• Setting the equations equal: \( f(x) = g(x) \).
• Here it was set as \( x^2 + 5 = 2x^2 + 1 \).

Rearranging this equation allows us to solve for the \( x\) values: \( x = -2 \) and \( x = 2 \). These values signify where the curves intersect and establish the bounds for integrating and calculating the area. Identifying these points is a foundational step, as it ensures that we know the exact region we need to consider to find the enclosed area.