Mathematics: 100PTS 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 :(

Pablo decides to save his money in a piggy bank and each day he saves $ 100 more than the day before. How much would he save on day 20 if he had $ 100 on the first day? Express the function and solveStep-by-step explanation:...

100PTS Sketch the region enclosed by the given

Ответ:

7rnmartinez

20.01.2022

Mathematics

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

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:

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

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100PTS 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 :(

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