Problem 32 - Calculus and its applications

Question

Find the area of the region bounded by the graphs of the given equations. $$ y=2 x^{2}-x-3, y=x^{2}+x $$

Step-by-Step Solution

Answer

The area between the curves is \( \frac{19}{3} \).

Step 1: Determine Points of Intersection

To find where the curves intersect, set the equations equal to each other: \[ 2x^2 - x - 3 = x^2 + x \] Rearranging gives us: \[ x^2 - 2x - 3 = 0 \] Factor this equation: \[ (x-3)(x+1) = 0 \] So, the points of intersection are \( x = 3 \) and \( x = -1 \).

Step 2: Set Up the Integral

The integral to find the area between the curves is given by: \[ \int_{-1}^{3} [(2x^2 - x - 3) - (x^2 + x)] \, dx \] Simplify the integrand: \[ \int_{-1}^{3} (x^2 - 2x - 3) \, dx \]

Step 3: Integrate the Function

Perform the integration to find the area: \[ \int (x^2 - 2x - 3) \, dx = \frac{x^3}{3} - x^2 - 3x + C \] We do not need \( C \) since we will calculate a definite integral.

Step 4: Evaluate the Definite Integral

Calculate the definite integral from \( x = -1 \) to \( x = 3 \): \[ \left[ \frac{x^3}{3} - x^2 - 3x \right]_{-1}^{3} \] Plug \( x = 3 \) and \( x = -1 \) into the antiderivative and subtract: \[ \left( \frac{27}{3} - 9 - 9 \right) - \left( \frac{-1}{3} - 1 + 3 \right) \] This simplifies to: \[ (9 - 9 - 9) - \left( -\frac{1}{3} - 1 + 3 \right) \] \[ -9 - \left( \frac{8}{3} \right) \] \[ = -9 + \frac{8}{3} \] Convert \(-9\) to thirds: \[ -\frac{27}{3} + \frac{8}{3} = -\frac{19}{3} \]

Step 5: Calculate the Absolute Area

Since we are finding the area, take the absolute value of the result from Step 4: \[ \left|-\frac{19}{3}\right| = \frac{19}{3} \] This is the area of the region bounded by the two curves.

Key Concepts

Area between curves

The concept of finding the area between curves is an important application of integration in calculus. When you have two curves on a graph, the area "between" them is just the space that's enclosed by these curves. To find this area, you'll typically need:

• The equations of the curves.
• The points where the curves intersect.

Finding the area is a way to measure the "size" of this space, which is technically called the "area bounded by the curves." This is often required in many calculus problems to understand the space defined by different functions. To find the area, you perform the following steps:

• First, determine where the curves intersect by setting their equations equal to each other. This will give the limits of integration.
• Next, set up a definite integral that represents the "top" curve's equation minus the "bottom" curve's equation between the points of intersection.
• Finally, evaluate the integral to find the area.

Definite integral

A definite integral is a critical concept in calculus used to find the net area under a curve within a given interval. It's called "definite" because it's evaluated over a specific range, unlike indefinite integrals that give a general form. The definite integral of a function is symbolized as \[ \int_{a}^{b} f(x) \, dx \] Here, \(a\) and \(b\) are the boundaries of integration, also known as the limits of integration. To find the definite integral:

• First, find the antiderivative of the function \(f(x)\) within the integral.
• Then, substitute the upper limit \(b\) and the lower limit \(a\) into this antiderivative.
• Subtract the value obtained from substituting \(a\) from the value obtained from \(b\).

In our context, the definite integral helps in calculating the area between two curves by integrating the difference of their equations over their points of intersection.

Points of intersection

Finding the points of intersection between curves is a crucial step in many calculus problems involving area. These points are where two functions cross each other, and they help in determining the limits for the integration interval when finding the area between curves.To find these points:

• Set the equations of the curves equal. This means solving \(f(x) = g(x)\).
• Rearrange this equation so that all terms are on one side, and simplify or factor if needed.
• Solving for \(x\) gives the x-values of the intersection points.

In the given problem, the points were found by setting \(2x^2 - x -3 = x^2 + x\), leading to the x-values \( x = -1 \) and \( x = 3 \). These are the boundaries for calculating the area between our given curves.