Calculate the area between two curves by integrating with respect to [latex]x[/latex]
Calculate the area of a compound region
Calculate the area between two curves by integrating with respect to [latex]y[/latex]
Determine the most effective variable, [latex]x[/latex] or [latex]y[/latex], for integration based on the curves’ orientation
Area of a Region between Two Curves
The Main Idea
Extends the concept of definite integrals to calculate areas of more complex regions
For continuous functions [latex]f(x)[/latex] and [latex]g(x)[/latex] where f(x) ≥ g(x) on [latex][a,b][/latex]:
[latex]A = \int_a^b [f(x) - g(x)] dx[/latex]
Area represents the difference between the upper and lower function, integrated over the interval
Process:
Identify the upper and lower functions
Determine the interval of integration
Set up and evaluate the definite integral
Often necessary to solve [latex]f(x) = g(x)[/latex] to determine integration limits
If [latex]R[/latex] is the region bounded by the graphs of the functions [latex]f(x)=\frac{x}{2}+5[/latex] and [latex]g(x)=x+\frac{1}{2}[/latex] over the interval [latex]\left[1,5\right],[/latex] find the area of region [latex]R.[/latex]
[latex]12[/latex] units2
If [latex]R[/latex] is the region bounded above by the graph of the function [latex]f(x)=x[/latex] and below by the graph of the function [latex]g(x)={x}^{4},[/latex] find the area of region [latex]R.[/latex]
[latex]\frac{3}{10}[/latex] unit2
Watch the following video to see the worked solution to this example.
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
Extends area calculations to regions where functions intersect
Key formula for regions bounded by intersecting curves [latex]f(x)[/latex] and [latex]g(x)[/latex] on [latex][a,b][/latex]:
[latex]A = \int_a^b |f(x) - g(x)| dx[/latex]
Absolute value in the formula accounts for changes in which function is greater
Process:
Identify intersection points of the functions
Break the interval into subintervals where one function is consistently greater
Set up and evaluate separate integrals for each subinterval
Sum the results to get the total area
If [latex]R[/latex] is the region between the graphs of the functions [latex]f(x)= \sin x[/latex] and [latex]g(x)= \cos x[/latex] over the interval [latex]\left[\frac{\pi}{2},2\pi \right],[/latex] find the area of region [latex]R.[/latex]
The two curves intersect at [latex]x=\frac{(5\pi )}{4}.[/latex]
[latex]2+2\sqrt{2}[/latex] units2
Watch the following video to see the worked solution to this example.
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
Consider the region depicted below. Find the area of [latex]R.[/latex]
Figure 6. Two integrals are required to calculate the area of this region.
We need to divide the interval into two pieces. The graphs of the functions intersect at [latex]x=1[/latex] (set [latex]f(x)=g(x)[/latex] and solve for [latex]x[/latex]), so we evaluate two separate integrals: one over the interval [latex]\left[0,1\right][/latex] and one over the interval [latex]\left[1,2\right].[/latex]
Over the interval [latex]\left[0,1\right],[/latex] the region is bounded above by [latex]f(x)={x}^{2}[/latex] and below by the [latex]x[/latex]-axis, so we have:
Over the interval [latex]\left[1,2\right],[/latex] the region is bounded above by [latex]g(x)=2-x[/latex] and below by the [latex]x\text{-axis,}[/latex] so we have:
Therefore, the area of the region is [latex]\frac{16\sqrt{2}}{3}[/latex] square units.
Regions Defined with Respect to [latex]y[/latex]
The Main Idea
Alternative approach to finding areas between curves by integrating with respect to [latex]y[/latex]
Useful when curves are more easily expressed as functions of [latex]y[/latex]
Can simplify calculations by requiring only one integral instead of two
Key formula: [latex]A = \int_c^d [u(y) - v(y)] dy[/latex], where [latex]u(y)[/latex] is the right boundary and [latex]v(y)[/latex] is the left boundary
Process:
Express curves as functions of [latex]y[/latex]
Determine the limits of integration ([latex]y[/latex]-coordinates of intersection points)
Set up and evaluate the integral
Find the area of the region bounded by [latex]y = x^2[/latex] and [latex]y = 2x[/latex].
Express curves as functions of [latex]y[/latex]:
[latex]x = \sqrt{y}[/latex] and [latex]x = \frac{y}{2}[/latex]
Find intersection points:
[latex]\sqrt{y} = \frac{y}{2}[/latex]
[latex]y = 0[/latex] or [latex]y = 4[/latex]