Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 10

Sketch the region enclosed by the curves $y = 1 + \sqrt{x}$, $\displaystyle y = \frac{(3+x)}{3}$. Then find the area of the region.


By using vertical strip,
$\displaystyle A = \int^{x_2}_{x_1} \left(y_{\text{upper}} - y_{\text{lower}} \right) dx$
In order to get the values of the upper and lower limits, we equate the two functions to get its points of intersection. Thus,

$
\begin{equation}
\begin{aligned}
1 + \sqrt{x} &= \frac{3+x}{3}\\
\\
3 + 3 \sqrt{x} &= 3+x\\
\\
3\sqrt{x} &= x \\
\\
9x &= x^2\\
\\
x^2 - 9x &= 0\\
\\
x(x-9) &= 0
\end{aligned}
\end{equation}
$

We have, $x = 0$ and $x =9$
Therefore,

$
\begin{equation}
\begin{aligned}
A &= \int^{9}_{0} \left[ (1+\sqrt{x}) - \left( \frac{3+x}{3} \right)\right] dx \\
\\
A &= \int^9_0 \left( \sqrt{x} - \frac{x}{3} \right) dx\\
\\
A &= \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^2}{2(3)} \right]^9_0\\
\\
A &= 4.5 \text{ square units}
\end{aligned}
\end{equation}
$

Comments

Post a Comment

Popular posts from this blog

Calculus: Early Transcendentals, Chapter 9, 9.3, Section 9.3, Problem 18

This differential equation can be solved by the method of separating variables. Multiply both sides by dt and divide both sides of the equation by L^2 . This will bring all terms depending on t to the right side and all terms depending on L to the left side: (dL)/L^2 = kln(t)dt Now both sides can be integrated. On the left side, we have an integral of a power function: int (dL)/L^2 = int L^(-2) dL = -1L^(-1) = -1/L (The constant of integration can be omitted for now.) On the right side, the integral of ln(t) equals int ln(t)dt = tln(t) - t (This can be verified by taking the derivative using the product rule and the fact that (ln(t))' = 1/t : (tln(t) - t)' = ln(t) + t/t - 1 = ln(t)) So the after the integration the equation becomes -1/L = k(tln(t) - t) + C The constant of integration C is added to the right side here. Since the initial condition is given, L(1) = -1, C can be found now: for t = 1, ln(t) = ln(1) = 0, and -1/(-1) = k(1*0 - 1) + C = -k + C 1 = -k + C, so from here...
Read more

In what ways might RFID technology be used to serve customers better? What problems might arise? Do you think that the technology might be valuable when implanted in animals or people?

RFID stands for Radio Frequency Identification. It is a type of wireless technology and basically has three components to it: a tag, a reader, and a computer system. The tag consists of a microchip and radio antenna. The microchip in the tag contains essential information about a product or item. To transmit this information to a reader, the tag uses radio signals. After picking up the signals, the reader delivers the information to a computer system. From the computer system, companies can easily track the kinds of products consumers like to buy. This allows companies to position advertisements, sales, and product placements in stores according to their customer's preferences. Also, using RFID technology increases productivity: it is much easier to use, is more accurate, and less error-prone than traditional bar-coding. Thus, increased productivity leads to greater profits for companies. As for consumers, the benefits of RFID technology are numerous. For example, it minimizes wait...
Read more

Single Variable Calculus, Chapter 7, 7.4-2, Section 7.4-2, Problem 52

Find the volume of the region under the curve $y = 10^{-x}$ from $x = 0$ to $x = 1$ that is rotated about the $x$-axis. By using vertical strips, notice that if you slice the figure, you'll get a cross section with radius $10^{-x}$, thus, the cross sectional area is $A = \pi (10^{-x})^2$. Hence, the volume is $ \begin{equation} \begin{aligned} V =& \int^1_0 A(x) dx \\ \\ V =& \int^1_0 \pi (10^{-x})^2 dx \\ \\ V =& \int^1_0 \pi \left( \frac{1}{10 x} \right)^2 dx \\ \\ V =& \pi \int^1_0 10^{-2x} dx \end{aligned} \end{equation} $ Let $u = -2x$, then $du = -2 dx$, so $\displaystyle dx = \frac{-du}{2}$. When $x = 0, u = 0$ and when $x = 1, u = -2$. Therefore, $ \begin{equation} \begin{aligned} V =& \pi \int^{-2}_0 10^u \cdot \frac{-du}{2} \\ \\ V =& \frac{- \pi}{2} \int^{-2}_0 10^u du \\ \\ V =& \frac{- \pi}{2} \left[ \frac{10^u}{\ln 10} \right]^{-2}_0 \\ \\ V =& \frac{- \pi}{2} \left[ \frac{10^{-2}}{\ln 10} - \frac{10^0}{\ln 10} \right] \\ \\ V =&...
Read more