The area represents the integral of the absolute difference between the two functions over a given interval.

Look for the x-coordinates (or y-coordinates if integrating with respect to y) where the curves intersect; these are the limits of integration.

The curve that is 'above' (or to the right, if integrating with respect to y) is subtracted from the curve that is 'below' (or to the left) within each interval.

Approximate the area by dividing it into rectangles or other simple shapes and summing their areas.

It indicates a value of x (or y) where the two functions have the same value, marking a potential change in which function is greater.

The area between the graph of $f(x) - g(x)$ and the x-axis represents the area between the curves $f(x)$ and $g(x)$.

Visually inspect the graph to see which function has greater y-values (or x-values if integrating with respect to y) within that interval.

If $f(x) - g(x) > 0$, then $f(x)$ is above $g(x)$; if $f(x) - g(x) < 0$, then $f(x)$ is below $g(x)$.

Look for points where the curves intersect; these points indicate where the 'top' and 'bottom' functions switch.

It shows that regardless of which function is on top, the area is always positive, as the absolute value ensures we're considering the magnitude of the difference.

1. Determine intersection points (0, 1). 2) Identify top/bottom functions (x^2 > x^3). 3) Integrate: $\int_{0}^{1} (x^2 - x^3) , dx$. 4) Evaluate: [x^3/3 - x^4/4]_0^1 = 1/12.

1. Find intersection points. 2) Graph functions. 3) Choose integration approach (dx or dy). 4) Set up integral(s) with correct intervals. 5) Evaluate integral(s).

Find the x-coordinates (or y-coordinates if integrating with respect to y) of the intersection points of the curves.

Split the integral into multiple integrals, each over an interval defined by consecutive intersection points, ensuring the correct order of subtraction within each interval.

Choose the variable that simplifies the integral setup; if the functions are easily expressed as functions of y, integrate with respect to y; otherwise, integrate with respect to x.

1. Find intersection points: $x^2 = 2x Rightarrow x = 0, 2$. 2) Split integral: $\int_{0}^{2} (2x - x^2) , dx + \int_{2}^{3} (x^2 - 2x) , dx$. 3) Evaluate each integral.

1. Find intersection points: $x = x^3 Rightarrow x = -1, 0, 1$. 2) Set up integrals: $\int_{-1}^{0} (x^3 - x) , dx + \int_{0}^{1} (x - x^3) , dx$. 3) Evaluate the integrals.

1. Find intersection points. 2) Sketch a graph (optional but recommended). 3) Determine which function is on top. 4) Set up and evaluate the definite integral.

1. Intersection points: $x^2 - 4 = 4 - x^2 Rightarrow x = -2, 2$. 2) Set up integral: $\int_{-2}^{2} (4 - x^2 - (x^2 - 4)) , dx$. 3) Evaluate the integral.

1. Express curves as x = f(y) and x = g(y). 2) Find intersection points in terms of y. 3) Integrate with respect to y: $\int_{c}^{d} |f(y) - g(y)| , dy$.

It involves integrating the absolute difference between two functions over an interval, accounting for changes in which function is 'on top'.

It ensures that the area is always positive, regardless of whether f(x) > g(x) or g(x) > f(x).

Intersection points define the intervals over which different functions act as the 'top' or 'bottom' curve, requiring separate integrals.

The definite integral calculates the signed area between a curve and the x-axis. Area between curves uses this to find the area between two functions.

It means that the function being integrated is below the x-axis, contributing a negative value to the integral unless absolute value is used.

Graphing helps visualize which function is on top and identify the intersection points, aiding in setting up the correct integrals.

Approximating the area by dividing it into thin vertical rectangles, where the height is the difference between the functions and the width is dx.

It's the method to find the x-coordinates where the two curves intersect, which are essential for defining the integration intervals.

The order determines whether the result is positive or negative; using absolute value corrects for this by ensuring a positive area.

The antiderivative is used to evaluate the definite integral at the bounds of integration, giving the net change over the interval.