Working Rules for Finding Area Using Definite Integrals

Working Rules for Finding Area Using Definite Integrals:

1. Positive Function:

• Rule 1: For a non-negative function $f\left(x\right)\ge 0$ over an interval $\left[a,b\right]$, the area under the curve $y=f\left(x\right)$ from $x=a$ to $x=b$ is given by the definite integral: $\text{Area}={\int }_{a}^{b}f\left(x\right)\text{\hspace{0.17em}}dx$
• This rule applies when the function remains above the x-axis for the entire interval, representing the area between the curve and the x-axis.
2. Multiple Regions and Piecewise Functions:

• Rule 2: If the function $f\left(x\right)$ changes sign or has multiple regions over $\left[a,b\right]$, find the areas of each region separately and sum their absolute values to get the total area.
• Here, $c$ is the point where $f\left(x\right)$ changes sign or the regions switch.
• Area between Curves:

• Rule 3: When finding the area between two curves $f\left(x\right)$ and $g\left(x\right)$ over $\left[a,b\right]$, compute the definite integral of their absolute difference: $\text{Area}={\int }_{a}^{b}\mid f\left(x\right)-g\left(x\right)\mid \text{\hspace{0.17em}}dx$
• This rule applies to determine the area enclosed by both curves within the given interval.
• Negative Function:

• Rule 4: If $f\left(x\right)\le 0$ over $\left[a,b\right]$, the definite integral ${\int }_{a}^{b}f\left(x\right)\text{\hspace{0.17em}}dx$ gives the negative of the area between the curve $y=f\left(x\right)$ and the x-axis.
• The magnitude of this integral represents the area, but the negative sign indicates that the area lies below the x-axis.
• Visualization and Graphical Analysis:

• Rule 5: Graphical representations help visualize the regions and understand the behavior of functions with respect to the x-axis for accurate area calculation using definite integrals.
• Riemann sums or numerical approximations aid in understanding and approximating areas under curves.

Example:

Consider the function $f\left(x\right)={x}^{2}-2x$ over the interval $\left[0,2\right]$. Let's find the area between the curve $y=f\left(x\right)$ and the x-axis within this interval.

1. Identify the Region:

• The function $f\left(x\right)={x}^{2}-2x$ intersects the x-axis at $x=0$ and $x=2$.
• To find the area between the curve and the x-axis, we'll compute the integral ${\int }_{0}^{2}\mid f\left(x\right)\mid \text{\hspace{0.17em}}dx$.
2. Integral Calculation: $\text{Area}={\int }_{0}^{2}\mid {x}^{2}-2x\mid \text{\hspace{0.17em}}dx$

• Factorizing: $\mid {x}^{2}-2x\mid =\mathrm{\mid }x\left(x-2\right)\mathrm{\mid }$
• Over $\left[0,2\right]$, $x\left(x-2\right)$ is positive, so $\mid {x}^{2}-2x\mid =x\left(x-2\right)$
• Integral Setup and Solution: $\text{Area}={\int }_{0}^{2}x\left(x-2\right)\text{\hspace{0.17em}}dx$

• Compute the integral: $\text{Area}={\left[\frac{{x}^{3}}{3}-{x}^{2}\right]}_{0}^{2}=\left[\frac{{2}^{3}}{3}-{2}^{2}\right]-\left[\frac{{0}^{3}}{3}-{0}^{2}\right]$$\text{Area}=\left[\frac{8}{3}-4\right]-\left[0-0\right]=\frac{8}{3}-4=\frac{8-12}{3}=-\frac{4}{3}$
• Result:

• The area between the curve $y={x}^{2}-2x$ and the x-axis over $\left[0,2\right]$ is $-\frac{4}{3}$ square units.
• The negative sign indicates that this area lies below the x-axis due to the nature of the function.