Understanding the Area Bounded by f(x), the x-axis, and Vertical Lines x=a and x=b

The concept of the area bounded by a function and the x-axis is a fundamental topic in integral calculus. It is often used to determine the region enclosed by the graph of a function, the x-axis, and two vertical lines x=a and x=b. This area can represent physical quantities like the distance traveled by an object under a velocity-time graph or the work done under a force-displacement graph.

Calculating the Area

The area ( A ) bounded by the curve ( y = f(x) ), the x-axis, and the vertical lines ( x = a ) and ( x = b ) is given by the definite integral:

[ A = \int_{a}^{b} |f(x)| \, dx ]

Here, ( |f(x)| ) is used to ensure that the area is always positive, as the integral of a function below the x-axis is negative.

Important Points:

• If ( f(x) \geq 0 ) for all ( x ) in ([a, b]), then the area is simply the integral of ( f(x) ) without the absolute value.
• If ( f(x) \leq 0 ) for all ( x ) in ([a, b]), then the area is the integral of ( -f(x) ) to make the area positive.
• If ( f(x) ) crosses the x-axis within the interval ([a, b]), the area must be calculated in segments where ( f(x) ) is entirely above or below the x-axis, and then the absolute values of these areas are summed.

Table of Differences and Important Points

Scenario	Description	Formula	Example
( f(x) \geq 0 )	The function is non-negative over ([a, b]).	( A = \int_{a}^{b} f(x) \, dx )	Area under ( y = x^2 ) from 0 to 1.
( f(x) \leq 0 )	The function is non-positive over ([a, b]).	( A = \int_{a}^{b} -f(x) \, dx )	Area above ( y = -x^2 ) from -1 to 0.
( f(x) ) crosses x-axis	The function changes sign over ([a, b]).	( A = \int_{a}^{c}	f(x)

Examples

Example 1: Non-negative Function

Consider the function ( f(x) = x^2 ) and the interval ([0, 1]). Since ( f(x) ) is non-negative over this interval, the area is:

[ A = \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{3} ]

Example 2: Non-positive Function

Consider the function ( f(x) = -x^2 ) and the interval ([-1, 0]). Since ( f(x) ) is non-positive over this interval, the area is:

[ A = \int_{-1}^{0} -(-x^2) \, dx = \int_{-1}^{0} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{0} = \frac{1}{3} ]

Example 3: Function Crossing the x-axis

Consider the function ( f(x) = x^3 - x ) and the interval ([-1, 1]). The function crosses the x-axis at ( x = -1, 0, ) and ( 1 ). We calculate the area in two parts:

[ A = \int_{-1}^{0} |x^3 - x| \, dx + \int_{0}^{1} |x^3 - x| \, dx ]

For ( x ) in ([-1, 0]), ( f(x) \leq 0 ), so we integrate ( -(x^3 - x) ). For ( x ) in ([0, 1]), ( f(x) \geq 0 ), so we integrate ( x^3 - x ):

[ A = \int_{-1}^{0} -(x^3 - x) \, dx + \int_{0}^{1} (x^3 - x) \, dx ] [ A = \left[ -\frac{x^4}{4} + \frac{x^2}{2} \right]{-1}^{0} + \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]{0}^{1} ] [ A = \left( 0 + \frac{1}{2} \right) - \left( \frac{1}{4} - \frac{1}{2} \right) + \left( \frac{1}{4} - \frac{1}{2} \right) ] [ A = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} ]

In conclusion, understanding the area bounded by a function and the x-axis involves determining whether the function is above or below the x-axis and calculating the definite integral accordingly. It is important to consider the sign of the function over the interval to ensure the area is positive.