I have to calculate area between the 3 curves: $$y=\frac1x\quad y=x\quad x=e$$ I integrated $(x-1/x$) from 1 to $e$, but it does not match any option. The given answer was $\frac32$.
Can you explain the the approach?
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$\begingroup$ The book is wrong. $\endgroup$– egregFeb 14 '18 at 10:36
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$\begingroup$ you mean 3/2 isn't the correct answer? $\endgroup$– user5722540Feb 14 '18 at 10:45
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$\begingroup$ Yes, the correct area is $\int_1^e(x-\frac{1}{x})\,dx$ if the text you reported is right: the three curves delimit a well determined bounded region that can be clearly seen in the picture in A.k.'s answer. $\endgroup$– egregFeb 14 '18 at 10:51
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If $y=1/x$ bounds the region from above and the $x$-axis is implied to bound the region from below, the given answer is obtained: $$\int_0^1x\,dx+\int_1^e\frac1x\,dx=\left[\frac{x^2}2\right]_0^1+[\ln x]_1^e=\left(\frac12-0\right)+(1-0)=\frac32$$ If $y=1/x$ bounds the region from below, leaving no implicit boundaries, a different answer is obtained: $$\int_1^e\left(x-\frac1x\right)\,dx=\left[\frac{x^2}2-\ln x\right]_1^e=\left(\frac{e^2}2-1\right)-\left(\frac12-0\right)=\frac{e^2-3}2$$ Good questions never leave anything to the reader to interpret, though. As egreg pointed out, the second answer has fewer assumptions and is the correct one.
answered Feb 14 '18 at 10:02
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$\begingroup$ I got the same answer, but the correct answer is not this. $\endgroup$ Feb 14 '18 at 10:04
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$\begingroup$ Can you please ignore my integral and rather confirm if you reach the same integral, as a fresh approach? $\endgroup$ Feb 14 '18 at 10:06
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$\begingroup$ should i include area from 0 to 1 also as suggested by other answer? $\endgroup$ Feb 14 '18 at 10:09
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$\begingroup$ 3/2 is the correct option but i am not convinced to use x axis as a boundary $\endgroup$ Feb 14 '18 at 10:11
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$\begingroup$ Leaving aside this quesiton, in general , there should not be any implied boundaries correct? $\endgroup$ Feb 14 '18 at 10:12
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you have 2 parts
from 0 to 1
&
from 1 to e
try to calculate the area between x and the x-axis
then between 1/x and x-axis
answered Feb 14 '18 at 10:02
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$\begingroup$ i think x axis is not to be included unless specified $\endgroup$ Feb 14 '18 at 10:09
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$\begingroup$ see the graph of the three curves $\endgroup$ Feb 14 '18 at 10:11
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$\begingroup$ you are also including x=0 as one of the curves, although it is not mentioned in the question $\endgroup$ Feb 14 '18 at 10:12
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$\begingroup$ Your picture clearly shows what is the area between the three curves, which isn't what you compute. $\endgroup$– egregFeb 14 '18 at 10:35