Question

100PTS!!! MATH HELP PLZ??!!
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.
y=1/x,y=1/x2,x=6

i tried to use the antiderivative but when i plug 0 in, it turns out ln(6)+1/6-ln(0)-1/0 and i dont know what to do with 1/0 :(

Answer 1

Answer:

$\displaystyle \int_1^6\left|\frac{1}{x}-\frac{1}{x^2}\right|dx=\ln|6|-\frac{5}{6}$

Step-by-step explanation:

We'll integrate with respect to x.

Recall that the area between some $f(x)$ and some $g(x)$ on the interval $[a, b]$ is given by:

$\displaystyle \int_a^b|f(x)-g(x)|dx$

The function $x=6$ will be the upper bounds of our definite integral. The lower bounds will be the intersection of $\displaystyle y=\frac{1}{x}$ and $\displaystyle y=\frac{1}{x^2}$. Set the functions equal to each other and solve for x:

$\displaystyle \frac{1}{x}=\frac{1}{x^2},\\\\x^2=x,\\x=1$

Therefore, we'll be integrating the area between $\displaystyle \frac{1}{x}$ and $\displaystyle \frac{1}{x^2}$ on the interval $[1, 6]$. In integral notation, this is:

$\displaystyle \int_1^6\left|\frac{1}{x}-\frac{1}{x^2}\right|dx$

To evaluate this integral, recall that $\displaystyle \int \frac{1}{x}dx=\ln|x|+C$ and $\displaystyle \int \frac{1}{x^2}dx=\int x^{-2}dx=-x^{-1}+C=-\frac{1}{x}+C$.

Therefore,

$\displaystyle \int_1^6\left|\frac{1}{x}-\frac{1}{x^2}\right|dx=\int_1^6\ln|x|-\left(-\frac{1}{x}\right)dx=\int_1^6\ln|x|+\frac{1}{x}dx$

Solving yields:

$\displaystyle\left \left(\ln|x|+\frac{1}{x}\right) \right \vert_1^6=\ln|6|+\frac{1}{6}-\left(\ln|1|+\frac{1}{1}\right)=\ln|6|+\frac{1}{6}-0-1=\boxed{\ln|6|-\frac{5}{6}}$

Answer 2

Answer:

$\ln 6 - \dfrac{5}{6}$

Step-by-step explanation:

To find the area enclosed by curves and lines, use definite integration:

$\displaystyle \int^b_a \text{f}(x)\:\:\text{d}x \quad \textsf{(where a is the lower limit and b is the upper limit)}$

Definite integrals have limits.  The limits tell you the range of x-values to integrate the function between.

Given functions:

$\begin{cases}y=\dfrac{1}{x}\\\\y=\dfrac{1}{x^2}\end{cases}$

The upper limit has been given as x = 6.

The lower limit is the point of intersection of the two curves.

To find the lower limit, equate the functions and solve for x:

$\implies \dfrac{1}{x}=\dfrac{1}{x^2}$

$\implies x=1$

Therefore, the limits for this integration are x = 1 and x = 6.

To find the area enclosed by the two curves, the point of intersection and the line x=3, integrate the areas under both curves using definite integration and subtract the area under the lower curve from the area under the upper curve.

$\begin{aligned}\displaystyle \int^6_1 \dfrac{1}{x}\:\:dx-\int^6_1 \dfrac{1}{x^2}\:\:dx & = \int^6_1\left(\dfrac{1}{x}-\dfrac{1}{x^2}\right)\:\:dx \\\\& = \int^6_1\left(\dfrac{1}{x}-x^{-2}\right)\:\:dx\\\\& = \left[\ln |x| + \dfrac{1}{x} \right]^6_1\\\\& = \left(\ln 6 + \dfrac{1}{6} \right) - \left(\ln 1 + \dfrac{1}{1} \right)\\\\& = \left(\ln 6 + \dfrac{1}{6} \right) - \left(0 +1\right)\\\\& = \ln 6 -\dfrac{5}{6}\end{aligned}$

Integration Rules

$\boxed{\begin{minipage}{4 cm}\underline{Integrating x^n }\\\\ \displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C} \end{minipage}}$

Increase the power by 1, then divide by the new power.

$\boxed{\begin{minipage}{3.5 cm}\underline{Integrating \frac{1}{x} }\\\\ \displaystyle \int \dfrac{1}{x}\:\text{d}x=\ln |x|+\text{C} \end{minipage}}$

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