# Topic 5 Applications of Integrals

## 5.1 Area between Curves

To find the area between curves, we identify a variable and use the differential in that variable as the width and the distance between the curve as the height and approximate the area using the Riemann sum of such a partition.

If the curves are given by \(y=f(x)\) and \(y=g(x)\) with \(a\le x\le b\), then the area between the curves may be set up as \[ \text{Area Between Curves}=\int_a^b|f(x)-g(x)|\mathrm{d}x \] Because the distance in \(y\) direction is easier to calculate.

If the curves are given by \(x=p(y)\) and \(x=q(y)\) with \(c\le y\le d\), then the area between the curves may be set up as \[ \text{Area Between Curves}=\int_c^d|p(y)-q(y)|\mathrm{d}y \] Because the distance in \(x\) direction is easier to calculate.

The absolute value sign can be removed by breaking the area into sub-areas.

**Exercise 5.1 **Sketch the graph of curves \(y=4-x^2\), \(y=2x+1\), \(x=-2\), and \(x=0\), and find the area enclosed by those curves.

*Solution*.

The graph of the curves shows that a vertical rectangular slice of the region has an area \([(4-x^2)-(2x+1)]\mathrm{d} x\).

Then the area of the region can be calculated by the definite integral \[ \int_{-2}^0[(4-x^2)-(2x+1)]\mathrm{d} x=(-\frac{x^3}{3}-x^2+3x)|_{-2}^0=\frac{22}{3}. \]

**Exercise 5.2 **Sketch the graph of curves \(y=x^2-2\) and \(y=2\), and find the area enclosed by those curves.

*Solution*.

Solve the system of equations \[ \begin{cases} y=x^2-2\\ y=2, \end{cases} \] we get two intersection points \((-2, 2)\) and \((2, 2)\). Over the interval \([-2, 2]\), \(x^2-2\le 2\). Then the area of the region can be calculated by the definite integral \[ \int_{-2}^2[2-(x^2-2)]\mathrm{d} x=2\int_{0}^2(4-x^2)\mathrm{d} x =2(4x-\frac{x^3}{3})|_{0}^2=\frac{32}{3}. \]

**Exercise 5.3 **Sketch the graph of curves \(y=x^2+x-2\) and \(y=1-x\), and find the area enclosed by those curves.

*Solution*.

Solve the system of equations \[ \begin{cases} y=x^2+x-2\\ y=1-x, \end{cases} \] we get two intersection points \((-3, 4)\) and \((1, 0)\). Since the function \((x^2+x-2)\le (1-x)\) for \(-3\le x\le 1\), the area of the region can be calculated by the definite integral \[ \int_{-3}^1[(1-x)-(x^2+x-2)]\mathrm{d} x=(3x-\frac{x^2}{2}-\frac{x^3}{3})|_{-3}^1=\frac{32}{3}. \]

**Exercise 5.4 **Sketch the graph of curves \(y=\cos x\) and \(y=2\sin x\cos x\) with \(-\pi/2\le x\le \pi/2\) and find the area between those curves.

*Solution*.

Solve the system of equations \[ \begin{cases} y=\cos x\\ y=2\sin x\cos x, \end{cases} \] we get three intersection points \((-\pi/2, 0)\), \((\pi/6, \sqrt{3}/2)\) and \((\pi/2, 0)\). The graph of the functions are shown below.

The area of the region can be calculated by the sum of two definite integral \[ \begin{aligned} &\int_{-\pi/2}^{\pi/2}|\cos x-2\sin x\cos x|\mathrm{d} x\\ =&\int_{-\pi/2}^{\pi/6}[\cos x-2\sin x\cos x]\mathrm{d} x+\int_{\pi/6}^{\pi/2}[2\sin x\cos x-\cos x]\mathrm{d} x\\ =&-\frac14(1-2\sin x)^2\big|_{-\pi/2}^{\pi/6}+\frac14(2\sin x-1)\big|_{\pi/6}^{\pi/2}\\ =&\frac94+\frac14\\ =&\frac52. \end{aligned} \]

**Exercise 5.5 **Sketch the graph of curves \(y=|x|\) and \(y=-x^2+2\) and find the area enclosed by those curves.

*Solution*.

Solve the system of equations \[ \begin{cases} y=|x|\\ y=-x^2+2, \end{cases} \] we get three intersection points \((-1, 1)\), and \((1, 1)\). The graph of the functions are shown below.

The area of the region can be calculated by the definite integral \[ \begin{aligned} &\int_{-1}^{1}||x|-(-x^2+2)|\mathrm{d} x\\ =&2\int_{0}^{1}||x|-(-x^2+2)|\mathrm{d} x\\ =&2\int_{0}^{1}(-x^2+2-x)\mathrm{d} x\\ =&2(-\frac{x^3}{3}-\frac{x^2}{2}+2x)\big|_0^1\\ =&\frac73. \end{aligned} \]

**Exercise 5.6 **Sketch the graph of curves \(x+y^2=5\) and \(y^2=x+3\) and find the area enclosed by those curves.

*Solution*.

Solve the system of equations \[ \begin{cases} x+y^2=5\\ y^2=x+3, \end{cases} \] we get three intersection points \((1, -2)\), and \((1, 2)\). The graph of the functions are shown below.

The area of the region can be calculated by the definite integral \[ \begin{aligned} &\int_{-2}^{2}|(5-y^2)-(y^2-3)|\mathrm{d} y\\ =&2\int_{0}^{2}(-2y^2+8)\mathrm{d} y\\ =&2(-\frac{2y^3}{3}+8y)\big|_0^2\\ =&\frac{64}{3}. \end{aligned} \]

**Exercise 5.7 **Sketch the graph of curves \(y=x\) and \(x=y^2-y-3\) and find the area enclosed by those curves.

*Solution*.

Solve the system of equations \[ \begin{cases} y=x\\ x=y^2-y-3, \end{cases} \] we get three intersection points \((-1, -1)\), and \((3, 3)\). The graph of the functions are shown below.

The area of the region can be calculated by the definite integral \[ \begin{aligned} &\int_{-1}^{3}|y-(y^2-y-3)|\mathrm{d} y\\ =&\int_{-1}^{3}(-y^2+2y+3)\mathrm{d} y\\ =&(-\frac{y^3}{3}+y^2+3y)\big|_{-1}^3\\ =&\frac{32}{3}. \end{aligned} \]

**Exercise 5.8 **Sketch the graph of curves \(x=y^4\) and \(y^2=2-x\) and find the area enclosed by those curves.

*Solution*.

Solve the system of equations \[ \begin{cases} x=y^4\\ y^2=2-x, \end{cases} \] we get three intersection points \((1, -1)\), and \((1, 1)\). The graph of the functions are shown below.

The area of the region can be calculated by the definite integral \[ \begin{aligned} &\int_{-1}^{1}|y^4-(2-y^2)|\mathrm{d} y\\ =&2\int_{0}^{1}(2-y^2-y^4)\mathrm{d} y\\ =&2(-\frac{y^5}{5}-\frac{y^3}{3}+2y)\big|_{0}^1\\ =&\frac{44}{15}. \end{aligned} \]

**Exercise 5.9 **Sketch the graph of curves \(y=x^2-1\) and \(y=x^3+x^2-x-1\) and find the area enclosed by those curves.

*Solution*.

Let \(D(x)=(x^2-1)-(x^3+x^2-x-1)=-x^3+x\). Solve \(D(x)=0\), we get three solutions \(x=-1\), \(x=0\), and \(x=1\). Using test point method or algebraic method, we find that \(D(x)<0\) for \(-1<x<0\) and \(D(x)>0\) for \(0<x<1\). The graph of the functions are shown below.

The area of the region can be calculated by the sum of two definite integral \[ \begin{aligned} &\int_{-1}^{1}|-x^3+x|\mathrm{d}x\\ =&2\int_{0}^{1}(-x^3+x)\mathrm{d} x =&2(-\frac{x^4}{4}+\frac{x^2}{2})\big|_{0}^1\\ =&\frac12. \end{aligned} \]

**Exercise 5.10 **Sketch the graph of curves \(x-2y=0\) and \(y^3-y^2=x\) and find the area enclosed by those curves.

*Solution*.

Since \(x\) is easier to solve, it is better to use the sum of areas horizontal rectangles, that is the area is given by \[ \int_a^b|x_R(y)-x_L(y)|\mathrm{d}y. \] Solve for \(x\) from the equations of the curves and set \(D(y)=(y^3-y^2)-2y=y^3-y^2-2y\). Solve \(D(y)=0\), we get three solutions \(y=-1\), \(y=0\), and \(y=2\). Using test point method or algebraic method, we find that \(D(y)>0\) for \(-1<y<0\) and \(D(y)<0\) for \(0<y<2\). The graph of the functions are shown below.

The area of the region can be calculated by the sum of two definite integral \[ \begin{aligned} &\int_{-1}^{2}|y^3-y^2-2y|\mathrm{d} y\\ =&\int_{-1}^{0}(y^3-y^2-2y)\mathrm{d} y-\int_{0}^{2}(y^3-y^2-2y)\mathrm{d} y\\ =&(\frac{y^4}{4}-\frac{x^3}{3}-y^2)\big|_{-1}^0-(\frac{y^4}{4}-\frac{x^3}{3}-y^2)\big|_{0}^2\\ =&\frac{5}{12}+\frac83\\ =&\frac{37}{12}. \end{aligned} \]