Homework 3

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Mos Kullathon
921425216

1. Find the area of the bounded region enclosed by $y = x^2 , x + y = 2$.

$$\begin{darray}{c} y = x^2 &\implies& f(x) = x^2 \\ x + y = 2 &\implies& g(x) = 2-x \end{darray}$$

Finding intersections of $f$ and $g$.
$$x^2 = 2 - x \\ x^2 + x - 2 = 0 \\ (x-1)(x+2) = 0 \\ \therefore x = -2, 1$$

Since $g(x) \ge f(x)$ for $x \in [-2, 1]$, the area between the curves is:
$$\begin{align*} \int_1^{-2} g(x)-f(x) \dd x &= \int_1^{-2} (2-x-x^2) \dd x \\ &= \bigg[2x - \frac{x^2}{2} - \frac{x^3}{3}\bigg]_1^{-2} \\ &= \frac{9}{2}. \end{align*}$$

2. Find the area of the region enclosed by $y=\tan x$ and the x-axis over the interval $[-\pi/3, \pi/4]$.

Since $\tan x \le 0$ for $x = 2\pi n + \pi$ where $n \in \Z$, we find that for values of $x\in [-\frac{\pi}{3}, \frac{\pi}{4}]$:

$$\tan x \le 0, \qquad x \in \[-\frac{\pi}{3}, 0\] \\ \tan x \ge 0, \qquad x\in \[0, \frac{\pi}{4}\]$$

As such, the total area of $y=\tan x$ bounded by the x-axis over $[-\frac{\pi}{3}, \frac{\pi}{4}]$ is:

$$\begin{align*} &-\int_{-\pi/3}^0 \tan x \dd x + \int_0^{\pi/4} \tan x \dd x \\ &= - \bigg[-\ln\cos x\bigg]_{-\pi/3}^0 + \bigg[-\ln\cos x\bigg]_0^{\pi/4} \\ &= \frac{3\ln 2}{2}. \end{align*}$$

3. Find the area of the shaded region.

$$\begin{darray}{c} x = (y-1)^2 &\implies& y = \sqrt x - 1 \\ x = 3 - y &\implies& y = 3 - x \\ x = 2\sqrt{y} &\implies& y = \frac{1}{4}x^2 \end{darray}$$

$\sqrt{x}-1 \ge \frac{1}{4}x^2$ for $x\in[0, 1]$. As such the area from $x=0$ to $x=1$ is:

$$\int_0^1 (\sqrt{x}+1-\frac{1}{4}x^2) \dd x.$$

$3-x \ge \frac{1}{4}x^2$ for $x\in[1, 2]$. As such the area from $x=1$ to $x=2$ is:

$$\int_1^2 (3-x-\frac{1}{4}x^2)\dd x.$$

Therefore, the area $A$ between the three functions from $x=0$ to $x=1$ is the sum of the two integrals.

$$\begin{align*} A &= \int_0^1 (\sqrt{x}+1-\frac{1}{4}x^2) \dd x + \int_1^2 (3-x-\frac{1}{4}x^2)\dd x \\ &= \bigg[\frac{2x^{3/2}}{3}+x-\frac{x^3}{12}\bigg]_0^1 + \bigg[3x-\frac{x^2}{2}-\frac{x^2}{12}\bigg]_1^2 \\ &= \frac{19}{12} + \frac{10}{3} - \frac{29}{12} \\ &= \frac{5}{2} \end{align*} \\$$

4. Book Section 2.1

(2) Area between $y=x^2$ and $y=3x+4$.

$$x^2 = 3x + 4 \\ x^2 - 3x - 4 = 0 \\ (x-4)(x+1) = 0 \\ \therefore x = -1, 4$$

$y=x^2$ and $y=3x+4$ intersects at $x = -1$ and $x = 4$, and $3x+4 \ge x^2$ for all $x\in[-1, 4]$. Therefore,

$$\begin{align*} A &= \int_{-1}^4 3x+4-x^2 \dd x \\ &= \[\frac{3x^2}{2}+4x-\frac{x^3}{3}\]_{-1}^4 \\ &= \frac{125}{6}. \end{align*}$$

(3) Area between $y=x^3$ and $y=x^2+x$.

$$x^3 = x^2 + x \\ x^3 - x^2 - x = 0 \\ \therefore x = \frac{1}{2}\pm\frac{\sqrt{5}}{2}$$

$y=x^3$ and $y=x^2+x$ intersects at $x = \frac{1-\sqrt{5}}{2}$ and $x=\frac{1+\sqrt{5}}{2}$.

$0 \ge x^3 \ge x^2 + x$ for all $x\in[\frac{1-\sqrt{5}}{2},0]$ and $x^2 + x \ge x^3 \ge 0$ for all $x\in[0,\frac{1+\sqrt{5}}{2}]$. Therefore,

$$\begin{align*} A &= \int_\frac{1-\sqrt{5}}{2}^0 (x^3 - x^2 - x) \dd x + \int_0^{\frac{1+\sqrt{5}}{2}} (x^2 + x - x^3) \dd x \\ &= \[\frac{x^4}{4} - \frac{x^3}{3} - \frac{x^2}{2}\]_\frac{1-\sqrt{5}}{2}^0 + \[\frac{x^3}{3} + \frac{x^2}{2} - \frac{x^4}{4}\]_0^{\frac{1+\sqrt{5}}{2}} \\ &= \frac{13}{12}. \end{align*}$$

(12) Area between $y=e$, $y=e^x$, and $y=e^{-x}$.

$$e = e^x \implies x = 1 \\ e = e^{-x} \implies x = -1 \\ \therefore x = \pm1$$

	$e = e^x$ at $x = 1$ and $\\e = e^{-x}$ at $x = -1$

$$\begin{align*} \displaystyle\therefore A &= \int_{-1}^0e-e^{-x}\dd x + \int_0^1 e-e^x\dd x \\ &= \bigg[ex+e^{-x}\bigg]_{-1}^0 + \bigg[ex - e^x\bigg]_0^1 \\ &= 2 \end{align*}$$

(22) Area between $x=-3+y^2$ and $x=y-y^2$.

$$-3+y^2 = y - y^2 \\ 2y^2-y-3 = 0 \\ \therefore y=-1,\frac{3}{2}$$

	$-3+y^2 = y - y^2\\$ at $y=-1$ and $\displaystyle y=\frac{3}{2}$

$$\begin{align*} \therefore A &= \int_{-1}^{3/2} y-y^2-(-3+y^2) \dd y \\ &= \int_{-1}^{3/2}-2y^2+y+3 \dd y \\ &= \bigg[-\frac{2y^3}{3} + \frac{y^2}{2} + 3y \bigg]_{-1}^{3/2} \\ &= \frac{125}{24} \end{align*}$$

(53) Area between $y^2 = x$ and $x=y+2$.

$$y^2 = y + 2 \\ y^2 - y - 2 = 0 \\ (y+1)(y-2) = 0 \\ \therefore y=-1,2$$

	$y^2=y+2\\$ at $\displaystyle y=-1$ and $y=2$

$$\begin{align*} \therefore A &= \int_{-1}^2 y+2-y^2 \dd y \\ &= \bigg[\frac{y^2}{2}+2y-\frac{y^3}{3}\bigg]_{-1}^2 \\ &= \frac{9}{2} \end{align*}$$