Objectives
- Use integrals to find area between curves
- Determine if a region should be integrated with respect to \(x\) or \(y\) in order to find the area between curves
Section 5.1 Areas Between Curves
Subsection 5.1.1 Before Class
https://mymedia.ou.edu/media/5.1-1/1_1i1v6tu0Subsubsection 5.1.1.1 Area
Example 5.1.1.
![The graph of \(f(x) = -(x-1)^2+3\) and \(g(x) = 1\) on the interval \([-1,3]\text{.}\)](generated/latex-image/image-139.svg)
- Write (and evaluate) an expression to find the area between \(f(x)\) and the \(x-\)axis on \([0,2]\text{.}\) Shade the area on the graph.
- Write (and evaluate) an expression to find the area between \(g(x)\) and the \(x-\)axis on \([0,2]\text{.}\) Shade the area on the graph.
- Write (and evaluate) an expression to find the area between \(f(x)\) and \(g(x)\) on \([0,2]\text{.}\) Do you need to shade the area?
Solution.
-
![This is the graph of \(f(x)\) on the interval \([0,2]\) with the area between the curve and the \(x-\)axis shaded](generated/latex-image/image-140.svg)
The area is found via the integral \(\ds \int_0^2 f(x)\, dx\text{,}\) and the area is \(\dfrac{16}{3}\text{.}\) -
![This is the graph of \(g(x)\) on the interval \([0,2]\) with the area between the curve and the \(x-\)axis shaded](generated/latex-image/image-141.svg)
The area is found via the integral \(\ds \int_0^2 g(x)\, dx\text{,}\) and the area is 2. -
![This is the graph of \(f(x)\) and \(g(x)\) on the interval \([0,2]\) with the area between the curves shaded](generated/latex-image/image-142.svg)
The area is found by combining the answers from (a) and (b): \(\ds \int_0^2 f(x)\, dx - \int_0^2 g(x)\, dx\text{,}\) so the area between the curves is \(\dfrac{10}{3}\text{.}\) We don’t necessarily need to shade the area, because if we did (a) and (b) on the same graph, the overlapping area is the area being excised.
Theorem 5.1.2. Area Between Curves.
Let \(f,g\) be continuous functions, and let \(f(x)\geq g(x)\) for all \(x\) in \([a,b]\text{.}\) The area \(A\) of the region bounded by the curves \(y=f(x),y=g(x)\) and the lines \(x=a,x=b\) is given by \(A = \ds \int_a^b (f(x)-g(x))\, dx\)
Example 5.1.3.
Find the area of the region bounded above by \(y = x^2+1\text{,}\) below by \(y = x\text{,}\) and on the sides by \(x = 0\) and \(x = 1\)
Solution.
\(\dfrac{5}{6}\)
Example 5.1.4.
Find the area of the region enclosed by the parabolas \(y = x^2\) and \(y = 2x-x^2\)
Solution.
\(\dfrac{1}{3}\)
Example 5.1.5.
Find the area of the region bounded by the curves \(y = \sin x\) and \(y = \cos x\text{,}\) from \(x = 0\) to \(x = \dfrac{\pi}{2}\)
Solution.
\(2\sqrt{2} - 2\)
Subsection 5.1.2 Pre-Class Activities
Example 5.1.6.
Find the area of the region enclosed by the curves \(y = x+1\) and \(y = 9-x^2\) between \(x = -1\) and \(x=2\text{.}\)
Solution.
\(\dfrac{39}{2}\)
Example 5.1.7.
Find the area of the region enclosed by the curves \(y = \sin x\) and \(y = x\) between \(x = \dfrac{\pi}{2}\) and \(x = \pi\text{.}\)
Solution.
\(\dfrac{3\pi^2}{8} - 1\)
Example 5.1.8.
Find the area between the curves \(y = (x-2)^2\) and \(y = x\text{.}\)
Solution.
\(\dfrac{9}{2}\)
Subsection 5.1.3 In Class
Subsubsection 5.1.3.1 Examples
Example 5.1.9.
Sketch the region enclosed by the curves \(y = \sqrt{x-1}\) and \(y=x-1\text{,}\) and find the area.
Solution.
![This is the graph of \(y = \sqrt{x-1}\) and \(y=x-1\) on the interval \([0,2.5]\) with the area between the curves shaded.](generated/latex-image/image-143.svg)
The area is \(\dfrac{7}{6}\text{.}\)
Example 5.1.10.
Sketch the region enclosed by the curves \(y = x^2\) and \(y = 4x-x^2\text{,}\) and find the area.
Solution.
![This is the graph of \(y = x^2\) and \(y = 4x-x^2\) on the interval \([-1,2.5]\) with the area between the curves shaded.](generated/latex-image/image-144.svg)
The area is \(\dfrac{8}{3}\)
Example 5.1.11.
Find the area enclosed by the line \(y = x-1\) and the parabola \(y^2=2x+6\)
Solution.
The area is 2
Example 5.1.12.
Sketch the region enclosed by the curves \(x = y^4\text{,}\) \(y=\sqrt{2-x}\text{,}\) and \(y = 0\text{,}\) and find the area.
Solution.
![This is the graph of \(x = y^4\text{,}\) \(y=\sqrt{2-x}\) on the interval \([0,2.5]\) with the area between the curves shaded.](generated/latex-image/image-145.svg)
The area is \(\dfrac{8}{3}\)
Example 5.1.13.
Find the area of the triangle with vertices \((2,0)\text{,}\) \((0,2)\text{,}\) and \((-1,1)\text{.}\)
Solution.
The area is 2
Example 5.1.14.
Sketch the region enclosed by the curves \(x = 2y^2\text{,}\) \(x = 4+y^2\text{,}\) and find the area.
Solution.
![This is the graph of \(x = 2y^2\) and \(x = 4+y^2\) on the interval \([-1,2.5]\) with the area between the curves shaded.](generated/latex-image/image-146.svg)
The shaded area is \(\dfrac{32}{3}\text{.}\)
Subsection 5.1.4 After Class Activities
Example 5.1.15.
Find the area of the region bounded by \(y = \sqrt{x}\text{,}\) \(y = x^{-2}\text{,}\) and \(x = 4\text{.}\)
Solution.
The area is \(\dfrac{47}{12}\)
Example 5.1.16.
Sketch the region bounded by the curves \(y = \cos \pi x\) and \(y = 4x^2 - 1\) and then find the area of the region.
Solution.
![This is the graph of the curves \(y = \cos \pi x\) and \(y = 4x^2 - 1\) on the interval \([-1,1]\) with the area between the curves shaded.](generated/latex-image/image-147.svg)
The area of the shaded region is \(\dfrac{2}{3} + \dfrac{2}{\pi}\)
Example 5.1.17.
Sketch the region enclosed by \(x = y - 1\text{,}\) \(y = 9-x^2\text{,}\) \(x = -1\text{,}\) and \(x = 2\text{,}\) then find the area of the region.
Solution.
![This is the graph of the curves \(x = y - 1\text{,}\) \(y = 9-x^2\text{,}\) \(x = -1\text{,}\) and \(x = 2\) on the interval \([-4,4]\) with the area between the curves shaded.](generated/latex-image/image-148.svg)
The area of the shaded region is \(\dfrac{39}{2}\)
Example 5.1.18.
Sketch the region enclosed by \(y = \dfrac{1}{x^2}\text{,}\) \(y = x\text{,}\) and \(y = \dfrac{1}{8}x\text{,}\) then find the area.
Solution.
![This is the graph of the curves \(y = \cos \pi x\) and \(y = 4x^2 - 1\) on the interval \([-1,1]\) with the area between the curves shaded.](generated/latex-image/image-149.svg)
The area of the shaded region is \(\dfrac{3}{4}\)
