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?

  • $\begingroup$ The book is wrong. $\endgroup$
    – egreg
    Feb 14 '18 at 10:36
  • $\begingroup$ you mean 3/2 isn't the correct answer? $\endgroup$ Feb 14 '18 at 10:45
  • $\begingroup$ 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. $\endgroup$
    – egreg
    Feb 14 '18 at 10:51

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.

  • $\begingroup$ I got the same answer, but the correct answer is not this. $\endgroup$ Feb 14 '18 at 10:04
  • $\begingroup$ 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
  • $\begingroup$ should i include area from 0 to 1 also as suggested by other answer? $\endgroup$ Feb 14 '18 at 10:09
  • $\begingroup$ 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
  • $\begingroup$ Leaving aside this quesiton, in general , there should not be any implied boundaries correct? $\endgroup$ Feb 14 '18 at 10:12

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

Graph of the Area

  • $\begingroup$ why 0 to 1 also? $\endgroup$ Feb 14 '18 at 10:05
  • $\begingroup$ i think x axis is not to be included unless specified $\endgroup$ Feb 14 '18 at 10:09
  • $\begingroup$ see the graph of the three curves $\endgroup$ Feb 14 '18 at 10:11
  • $\begingroup$ 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
  • $\begingroup$ Your picture clearly shows what is the area between the three curves, which isn't what you compute. $\endgroup$
    – egreg
    Feb 14 '18 at 10:35

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