Review 5: Areas Between Curves

Recall that the integral of a function LaTeX: f(x) $f(x)$ gives the area between the curve $f(x)$ and the LaTeX: x $x$ -axis. Using this we can determine the area of the following region:

$The graph of $y=x+1$ and $y=x{{\mathbf{e}}^{-{{x}^{2}}}}$ on the domain 0<x<2. The two graphs never intersect on this domain and the graph of $y=x+1$ is always larger than $y=x{{\mathbf{e}}^{-{{x}^{2}}}}$. The area between the two graphs has been shaded in.$

The area between the blue curve and the LaTeX: x $x$ -axis is given by $LaTeX: \int_0^2 x+1 \, dx=\tfrac12 x^2+x\Big|_0^2=2+2-0=4.$ $\int_0^2 x+1 \, dx=\tfrac12 x^2+x\Big|_0^2=2+2-0=4.$ and the area under the red curve is given by $LaTeX: \int_0^2 xe^{-x^2}\, dx=\int_{x=0}^{x=2} -\tfrac12 e^u \, du=-\tfrac12 e^{-x^2} \Big|_0^2=-\tfrac12 e^{-4}+\tfrac12.$ $\int_0^2 xe^{-x^2}\, dx=\int_{x=0}^{x=2} -\tfrac12 e^u \, du=-\tfrac12 e^{-x^2} \Big|_0^2=-\tfrac12 e^{-4}+\tfrac12.$ Thus the area between the two curves in the difference between these two integrals; thus the area is

$LaTeX: \begin{align*} \int_0^2 x+1 \, dx -\int_0^2 xe^{-x^2} \, dx&=4+\tfrac{e^{-4}}{2}-\tfrac12 . \end{align*}$ $\begin{align*} \int_0^2 x+1 \, dx -\int_0^2 xe^{-x^2} \, dx&=4+\tfrac{e^{-4}}{2}-\tfrac12 . \end{align*}$

We can apply this idea to find the area of more complicated regions as well. Consider the following region:

$The graph of $x=y+1$ and $x=\frac{1}{2}{{y}^{2}}-3$ on the domain -4<y<4.5. The two graphs intersect at y=-2 and y=4. In the domain -2<y<4 the graph of the line is larger (i.e. to the right) than the graph of the parabola. This is the region we are interested in for this problem.$

We wish to find the area enclosed by these two curves. First, we need to find the intersection points between these two curves. These two curves intersect when $LaTeX: y+1=\tfrac12 y^2-3$ $y+1=\tfrac12 y^2-3$ . Solving for $y$ ,

$LaTeX: \begin{align*} y+1&=\tfrac12y^2-3 \\ 0&=\tfrac12 y^2-y-4 \\ 0&=\tfrac12 \big( y^2-2y-8\big) \\ 0&=\tfrac12(y-4)(y+2). \end{align*}$ $\begin{align*} y+1&=\tfrac12y^2-3 \\ 0&=\tfrac12 y^2-y-4 \\ 0&=\tfrac12 \big( y^2-2y-8\big) \\ 0&=\tfrac12(y-4)(y+2). \end{align*}$

Thus these curves intersect when LaTeX: y=4 $y=4$ and LaTeX: y=-2 $y=-2$ . Thus the coordinates for the intersection points are LaTeX: (5,4) $(5,4)$ and LaTeX: (-1,-2) $(-1,-2)$ . Now that we know these points, we have a decision to make: which variable to integrate in.

Approach 1) integrating in $x$ :

First, to integrate in LaTeX: x $x$ we need to make our curves functions of $x$ . Hence the parabola $LaTeX: x=\tfrac12 y^2-3$ $x=\tfrac12 y^2-3$ becomes two functions of $x$ , $LaTeX: y=\pm\sqrt{2x+6}$ $y=\pm\sqrt{2x+6}$ . Next we need to subtract the lower function from the higher function, but there are two different lower functions, first the lower half of the parabola, and second the line LaTeX: y=x-1 $y=x-1$ . A sketch of this split is shown below.

$The graph of $x=y+1$ and $x=\frac{1}{2}{{y}^{2}}-3$ on the domain -4<y<4.5. The two graphs intersect at y=-2 and y=4. Unlike the first graph in this example the graphs are done assuming they are functions of x. So, the line is graphed as $y=x-1$ and the top of the parabola is graphed as $y=\sqrt{2x+6}$ and the bottom of the parabola is graphed as $y=-\sqrt{2x+6}$. The area between the curves is shaded but is shaded based on the functions being functions of x instead of functions of y. So, in the range -3<x<-1 we get one shading because the top/bottom of the graph are the top/bottom “branches” of the parabola. In the range -1<x<5 we get a different shading because the top graph is the top “branch” of the parabola and the bottom graph is the line.$

Thus to find the area of the enclosed region we compute two different integrals,

$LaTeX: \int_{-3}^{-1} \sqrt{2x+6}-(-\sqrt{2x+6}) \, dx$ $\int_{-3}^{-1} \sqrt{2x+6}-(-\sqrt{2x+6}) \, dx$ and $LaTeX: \int_{-1}^5 \sqrt{2x+6}-(x-1) \, dx$ $\int_{-1}^5 \sqrt{2x+6}-(x-1) \, dx$ .

Adding these together we get

$LaTeX: \begin{align*} A&=\int_{-3}^{-1} 2\sqrt{2x+6} \, dx+\int_{-1}^5 \sqrt{2x+6}-x+1 \, dx \\ &=\Big[\tfrac23(2x+6)^{3/2}\Big]_{-3}^{-1}+\Big[\tfrac13(2x+6)^{3/2}-\tfrac12x^2+x\Big]_{-1}^5 \\ &=\tfrac23(4)^{3/2}-0+\tfrac13(16)^{3/2}-\tfrac{25}2+5-\tfrac13(4)^{3/2}+\tfrac12+1 \\ &=\tfrac{8}{3}+\tfrac{64}{3}-6 \\ &=18. \end{align*}$ $\begin{align*} A&=\int_{-3}^{-1} 2\sqrt{2x+6} \, dx+\int_{-1}^5 \sqrt{2x+6}-x+1 \, dx \\ &=\Big[\tfrac23(2x+6)^{3/2}\Big]_{-3}^{-1}+\Big[\tfrac13(2x+6)^{3/2}-\tfrac12x^2+x\Big]_{-1}^5 \\ &=\tfrac23(4)^{3/2}-0+\tfrac13(16)^{3/2}-\tfrac{25}2+5-\tfrac13(4)^{3/2}+\tfrac12+1 \\ &=\tfrac{8}{3}+\tfrac{64}{3}-6 \\ &=18. \end{align*}$

Approach 2) integrating in $y$ :

We can write both curves as functions of y, specifically $LaTeX: x=\tfrac12y^2-3$ $x=\tfrac12y^2-3$ and LaTeX: x=y+1 $x=y+1$ . As in the case of integrating in $x$ , we subtract the smaller function from the larger function, then integrate. If integrating in LaTeX: y $y$ , visually that means subtracting the leftmost function from the rightmost function.

$The graph of $x=y+1$ and $x=\frac{1}{2}{{y}^{2}}-3$ on the domain -4<y<4.5. The two graphs intersect at y=-2 and y=4. In this graph the area between the two graphs are shaded given that the functions are functions of y. In this case the line is always on the right and the parabola is always on the left.$

Since there is only one rightmost and one leftmost function within the bounds of the region, our integral is much simpler. The area is given by

$LaTeX: \begin{align*} A&=\int_{-2}^4 y+1-(\tfrac12y^2-3) \, dy \\ &=\tfrac12y^2+y-\tfrac16 y^3+3y\Big|_{-2}^4 \\ &=8+4-\tfrac{64}{6}+12-\Big(2-2+\tfrac86-6\Big) \\ &=18. \end{align*}$ $\begin{align*} A&=\int_{-2}^4 y+1-(\tfrac12y^2-3) \, dy \\ &=\tfrac12y^2+y-\tfrac16 y^3+3y\Big|_{-2}^4 \\ &=8+4-\tfrac{64}{6}+12-\Big(2-2+\tfrac86-6\Big) \\ &=18. \end{align*}$

Notice that this approach is much simpler than integrating in LaTeX: x $x$ and gives the same area.