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$\ln x$ intersects the y axis at $-\infty$

So $$A=\int_{-\infty}^{e} \ln x \,dx-\int_{-\infty}^{e} \frac xe dx$$

Which isn’t really a proper definite integral since I am having trouble applying the limits

What is the correct way to do it?

Aditya
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    I don't understand $\int_{-\infty}^e\ln x,dx$: the domain of $\ln$ is $(0\infty)$. –  Dec 20 '20 at 11:32

3 Answers3

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Your lower bounds should be $0$ since the $y$-axis is stated as a bound. Then the integral is $$\int_0^e(x/e-\ln x)\,dx=[x^2/2e-x\ln x+x]_0^e=\frac e2$$

Parcly Taxel
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  • I'd worry about knowing that, (as you have an improper integral) as the lower limit tends towards zero, $xlnx$ tends to zero. You have assumed it does. Luckily it does but to prove it, the easiest way would be via L'Hospital's Rule. I mention this because the question indicates that limits should be being applied. – Martin Hansen Dec 20 '20 at 12:30
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Notice that due to $\ln x$ being one of the function, an immediate constraint is $ x>0$ , so we only have to consider half of the $x-y$ plane.

Sketching the graph:

enter image description here

Legend: Red = ln(x), blue = $ \frac{x}{e}$

Observe both curves intersect at x=e.

Hence, we can by inspection write down area under curve as:

$$ |\int_0^1 \ln x dx| + \int_0^1 x dx + \int_1^e ( x - \ln x ) dx$$

tryst with freedom
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enter image description here

Considering the graphs of $y = \frac x e$ and $y = \ln x$, the bounds should be $(0,e)$, and $\frac x e > \ln x$. Hence you are looking for the integral:

$$\int_0^e \left(\frac xe - \ln x\right) \ dx$$

player3236
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