Solutions to Problem Set 10

Math 211-03

9-25-2017

[Area]

1. Find the area of the region between $y = x^2 + 1$ and $y = x$ from $x = -1$ to $x = 1$ .

$$\hbox{\epsfysize=1.5in \epsffile{area7.eps}}$$

The curves do not intersect on the interval $-1 \le x \le 1$ .

The top curve is $y = x^2 + 1$ and the bottom curve is $y = x$ . The area is

$$\int_{-1}^1 [(x^2 + 1) - x]\,dx = \int_{-1}^1 (x^2 + 1 - x)\,dx = \left[\dfrac{1}{3} x^3 + x - \dfrac{1}{2} x^2\right]_{-1}^1 = \dfrac{8}{3} = 2.66666 \ldots.\quad\halmos$$


2. Find the area of the region between $y = x^2 + 1$ and $y = -5 -
   x$ from $x = -1$ to $x = 2$ .

$$\hbox{\epsfysize=1.75in \epsffile{area1.eps}}$$

The curves do not intersect on the interval $-1 \le x \le 2$ . Using vertical rectangles, I get the following integral for the area:

$$\int_{-1}^2 \left((x^2 + 1) - (-5 - x)\right)\,dx = \int_{-1}^2 \left(x^2 + x + 6\right)\,dx = \left[\dfrac{1}{3}x^3 + \dfrac{1}{2}x^2 + 6 x\right]_{-1}^2 = \dfrac{45}{2} = 22.5.\quad\halmos$$


3. Find the area of the region bounded by the curves $y = x^2 - 16$ and $y = 6 x$ .

$$\hbox{\epsfysize=1.75in \epsffile{area2.eps}}$$

Set the curves equal to find the intersection points:

$$x^2 - 16 = 6 x, \quad x^2 - 6 x - 16 = 0, \quad (x - 8)(x + 2) = 0, \quad x = -2 \quad\hbox{or}\quad x = 8.$$

Using vertical rectangles, I get the following integral for the area:

$$\int_{-2}^8 \left(6 x - (x^2 - 16)\right)\,dx = \int_{-2}^8 \left(6 x - x^2 + 16)\right)\,dx = \left[3 x^2 - \dfrac{1}{3} x^3 + 16 x\right]_{-2}^8 = \dfrac{500}{3} \approx 166.66667.\quad\halmos$$


4. Find the area of the region bounded by the curves $y = 2 - x^2$ and $y = x^2 - 4
   x - 28$ .

$$\hbox{\epsfysize=1.5in \epsffile{area8.eps}}$$

Find the intersection points:

$$\eqalign{ 2 - x^2 & = x^2 - 4 x - 28 \cr 0 & = 2 x^2 - 4 x - 30 \cr 0 & = x^2 - 2 x - 15 \cr 0 & = (x - 5)(x + 3) \cr}$$

The curves intersect at $x =
   -3$ and at $x = 5$ . Since $y = 2 -
   x^2$ is a parabola opening downward and $y = x^2 - 4 x -
   28$ is a parabola opening upward, $y = 2 - x^2$ is the top curve and $y = x^2 - 4 x - 28$ is the bottom curve. The area is

$$\int_{-3}^5 [(2 - x^2) - (x^2 - 4 x - 28)]\,dx = \int_{-3}^5 (30 + 4 x - 2 x^2)\,dx =$$

$$\left[30 x + 2 x^2 - \dfrac{2}{3} x^3\right]_{-3}^5 = \dfrac{512}{3} = 170.66666 \ldots.\quad\halmos$$


5. Find the area of the region under $y = \dfrac{1}{(x + 3)^2}$ and above the x-axis, from $x =
   0$ to $+\infty$ .

$$\hbox{\epsfysize=1.75in \epsffile{area3.eps}}$$

The area is given by the following improper integral:

$$\int_0^\infty \dfrac{dx}{(x + 3)^2} = \lim_{b\to \infty} \int_0^b \dfrac{dx}{(x + 3)^2} = \lim_{b\to \infty} \left[-\dfrac{1}{x + 3}\right]_0^b = \lim_{b\to \infty} \left(-\dfrac{1}{b + 3} + \dfrac{1}{3}\right) = 0 + \dfrac{1}{3} = \dfrac{1}{3}.\quad\halmos$$


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