Hello I'm trying to solve a calculus problem, but I can't figure out the answer. The problem says: Find the area of the region bounded by the graphs of $y=\ln x$ and $y=x-5$ with respect to $x$ (part A) and with respect to $y$ (part B). I know that the first step is setting $\ln x=x-5$ to find the bounds for the definite integral. However, I only get one solution, $x=6.397$. I'm not sure what to do after this point.

  • $\begingroup$ Hint: draw the functions, and go from there. $\endgroup$ – user160069 Mar 7 '17 at 3:57
  • $\begingroup$ There is another solution near zero. Graph the two functions to see this must be the case. So maybe try Newton's method but with starting value near zero to get the second solution. $\endgroup$ – Ahmed S. Attaalla Mar 7 '17 at 3:59
  • $\begingroup$ Type ln(x)=x-5 into Wolfram Alpha, or go to this link: wolframalpha.com/input/?i=ln(x)%3Dx-5 $\endgroup$ – Arby Mar 7 '17 at 4:08
  • 1
    $\begingroup$ Thank you. I found x=.0068 and the final answer is 17.13 $\endgroup$ – Jwa8129 Mar 7 '17 at 4:27

Let me show that you have two intersections considering the function $$f(x)=\log(x)-x+5$$ $$f'(x)=\frac 1x-1$$ $$f''(x)=-\frac 1{x^2}$$ The first derivative cancels when $x=1$ and $f(1)=4$ which is a maximum (by the second derivative test).

So, there is one root $x_1<1$ and a second $x_2>1$. Tou can easily compute tehm using a numerical method such as Newton.

Just for your curiosity, let me add that, sooner or later, you will learn taht any equation which can write $$A+Bx+C\log(D+Ex)=0$$ has solutions in terms of Lambert function. For this case,

$$x_1=-W\left(-\frac{1}{e^5}\right)\qquad x_2=-W_{-1}\left(-\frac{1}{e^5}\right)$$ The Wikipedia page gives series expansions which allow good approximations of the numerical values.

This makes that there is an analytical expression for the area you need to compute.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.