Answer
See the explanation
Work Step by Step
The area of the region between the graphs of two continuous functions can be calculated using definite integrals. If we have two functions, f(x) and g(x), and we want to find the area between their graphs over a specified interval [a, b], the formula for the area (A) is given by:
\[ A = \int_{a}^{b} |f(x) - g(x)| \, dx \]
In words, this formula calculates the integral of the absolute difference between the two functions over the specified interval. The absolute value ensures that the area is always positive.
Here's a simple example to illustrate this concept:
Consider two functions, \( f(x) = x^2 \) and \( g(x) = 2x - 1 \), and we want to find the area between their graphs over the interval [0, 2].
\[ A = \int_{0}^{2} |x^2 - (2x - 1)| \, dx \]
First, find the points of intersection by setting \( f(x) = g(x) \):
\[ x^2 = 2x - 1 \]
Solving for x, we get \( x = 1 \).
Now, break the integral into two parts over the interval [0, 1] and [1, 2]:
\[ A = \int_{0}^{1} (2x - 1 - x^2) \, dx + \int_{1}^{2} (x^2 - (2x - 1)) \, dx \]Evaluating the Integrals
For the first integral:
\[ \int_{0}^{1} (2x - 1 - x^2) \, dx = \left[ x^2 - \frac{1}{3}x^3 - x^3 \right]_{0}^{1} \]
\[ = \left(1 - \frac{1}{3} - 1\right) - \left(0 - 0 - 0\right) = \frac{1}{3} \]For the second integral:
\[ \int_{1}^{2} (x^2 - (2x - 1)) \, dx = \left[ \frac{1}{3}x^3 - (x^2 - 2x) \right]_{1}^{2} \]
\[ = \left(\frac{8}{3} - (4 - 4)\right) - \left(\frac{1}{3} - (1 - 2)\right) = \frac{1}{3} \]Add the results of the two integrals:
\[ A = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \]
So, the area between the graphs of \( f(x) = x^2 \) and \( g(x) = 2x - 1 \) over the interval \([0, 2]\) is \( \frac{2}{3} \) square units.
