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Integral Calculus lecture notes: Areas Between Curves
Eugene Kritchevski, Vanier College, last updated on September 25, 2017
We discuss how to compute the area of a region encosed by curves in a plane. We will see that the area can be found by computing a definite integral where the variable of integration is either $x$ or $y$. The challenge in such area computation problems is to understand the geometry and to set up correctly the area integral.
Integration with respect to $x$
Suppose that the region $\cal R$ lies between $y=y_B(x)$, $y=y_T(x)$ and the vertical lines $x=a$ and $x=b$, that is
$${\cal R}=\set{(x,y): a \leq x \leq b \textrm{ and } y_B(x) \leq y \leq y_T(x)}.$$
Here $y=y_T(x)$ is the top curve and $y=y_B(x)$ is the bottom curve. We assume that $y_B(x) \leq y_T(x)$ for all $x$ in $[a,b]$. In order to understand how to compute the area of the region $\cal R$, we approximate $\cal R$ by thin vertical rectangles. The figure below illustrates the contsruction.
The area of the region
$$\set{(x,y): a \leq x \leq b \textrm{ and } y_B(x) \leq y \leq y_T(x)},$$
where $y_T$ and $y_B$ are continuous and $y_T(x)\geq y_B(x)$ for all $x$ in $[a,b]$, is
$$A=\int_a^b [y_T(x)-y_B(x)]\,dx.$$
Note that Riemann sums are only used to justify the formula for the area. The actual computation of the area is usually done by evaluating the integral with FTC2.
Example 1 - direct application of the formula: The area bounded by $y=x+3$, $y=1/(1+x^2)$, $x=-1$ and $x=2$. Here the top curve is $y=x+3$ and the bottom curve is $y=1/(1+x^2)$. The exact area is then
$$A=\int_{-1}^{2}[x+3-\frac{1}{1+x^2}]\,dx=\of{\frac{x^2}{2}+3x-\arctan(x)}\Big|_{-1}^{2}=\of{2+6-\arctan(2)}-\of{\frac{1}{2}-3-\arctan(-1)}\simeq 8.607$$
Example 2 - we also have to compute the limits of integration: Let us find the area enclosed by the parabolas $y=x^2-6x+8$ and $y=4x-x^2$. Here we need to find the points of intersection of the two parabolas and to draw a picture. For the points of intersection, we set
$$x^2-6x+8=4x-x^2$$
which we can solve:
$$(x^2-6x+8)-(4x-x^2)=0$$
$$2x^2-10x+8=0$$
$$x^2-5x+4=0$$
$$(x-1)(x-4)=0$$
The solutions to the quadratic equation are $x_1=1$ and $x_2=4$. The corresponding $y$ values are $y_1=3$ and $y_2=0$ and so the two parabolas intersect at $(1,3)$ and $(4,0)$. On the interval $1\leq x \leq 4 $, the concave down parabola $y=4x-x^2$ is the top curve and the concave up parabola $y=x^2-6x+8$ is the bottom curve. The are of the enclosed region is thus given by
$$A=\int_{1}^{4}[(4x-x^2)-(x^2-6x+8)]\,dx=\int_{1}^{4}[-2 x^2 +10x-8]\,dx=-2\int_{1}^{4}[x^2 -5x+4]\,dx=-2\of{x^3/3-5x^2/2+4x}\Big |_{1}^{4}=\cdots=9$$
Integration with respect to $y$
Instead of integrating with respect to $x$, it is someimes more natural and easier to find the area of a region by integration with respect to $y$. In order to derive the formula, we approximate the region by thin rectangles parallel to the $x$-axis as in the figure below.
The area of the region
$$\set{(x,y): c \leq y \leq d \textrm{ and } x_L(y) \leq x \leq x_R(y)},$$
where $x_R$ and $x_L$ are continuous and $x_R(y)\geq x_L(y)$ for all $y$ in $[c,c]$, is
$\cal R$ is
$$A=\int_c^d [x_R(y)-x_L(y)]\,dy.$$
Example 3 - integration with repect to $y$ is easier than integration with respect to $x$: We want to compute the area of region enclosed by the parabola $x=y^2-1$ and the line $y=-x+1$. We isolate $x$ in the equation of the line: $x=1-y$. Then we find the points of intersection:
$$y^2-1=1-y$$
$$y^2+y-2=0$$
$$(y+2)(y-1)=0$$
This gives the solutions $y_1=-2$ and $y+2=1$. The correspondng $x$-values are $x_1=2$ and $x_2=0$. The intersection points are then $(3,-2)$ and $(0,1)$. On the interval $-2 \leq y \leq 1$, we have $x_R(y)=1-y$ and $x_L(y)=y^2-1$. The area is then
$$A=\int_{-2}^{1}[x_R(y)-x_L(y)]\,dy=\int_{-2}^{1}[(1-y)-(y^2-1))]\,dy=\int_{-2}^{1}[2-y-y^2]\,dy=\of{2y-y^2/2-y^3/3}\Big|_{-2}^{1}=\cdots=\frac{9}{2}.$$
The area can also be computed integrating with respect to $x$, but the computation is more complicated. We have
$$x=y^2-1 \Leftrightarrow y\pm\sqrt{x+1}.$$
On the interval $-1\leq x \leq 0$, the top curve is $y=\sqrt{x+1}$ and the bottom curve is $y=-\sqrt{x+1}$. On the interval $0 \leq x \leq 3$, the top curve is $y=-x+1$ and the bottom curve is $y=-\sqrt{x+1}$. The area integral has to be splitted:
$$A=\int_{-1}^{0}[(\sqrt{x+1})-(-\sqrt{x+1})]\,dx +\int_{0}^{3}[(-x+1))-(-\sqrt{x+1})]\,dx=\int_{-1}^{0}2 \sqrt{x+1}\,dx+\int_{0}^{3}[-x+1+\sqrt{x+1}]\,dx=\cdots=\frac{4}{3}+\frac{19}{6}=\frac{9}{2}$$
Exercises
Find the area of the region bounded by:
- $y=\sin(x)$, $y=\cos(x)$, $x=\pi/4$, and $x=5\pi/4$.
- $y=x^2$ and $y=x+2$.
- $y=2\sqrt{x}$, $y=0$, and $y=3\sqrt{x-5}$.
- $y=x^2$ and $y=x^5$.
- $y=3(1-x^2)$ and $y=4(1-x^2)^2$.
You can find the solution to each problem in
this video on areas between curves by Selwin Hollis.