Question

Which of the following integrals represents the area of the region bounded by x = e and the functions f(x) = ln(x) and g(x) = log1/e(x)?
the integral from 0 to e of the quantity, natural log of x minus the log base 1 over e of x, dx

the integral from 1 to e of the quantity, the log base 1 over e of x minus the natural log of x, dx

the integral from negative 1 to 0 of 4 minus the log base 1 over e of x,dx plus the integral from 0 to 1 of 4 minus the natural log of x, dx

the integral from 1 to e of the quantity, natural log of x minus the log base 1 over e of x, dx

Answer 1

Notice that

$\log_{1/e}x=\dfrac{\ln x}{\ln\frac1e}=\dfrac{\ln x}{-\ln e}=-\ln x$

$f(x)=\ln x$ and $g(x)=-\ln x$ intersect when $x=1$. For all $x>1$, we have $\ln x>0$ and $-\ln x$, so $f(x)>g(x)$. Then the area we want is given by the integral,

$\displaystyle\int_1^e\ln x-(-\ln x)\,\mathrm dx=2\int_1^e\ln x\,\mathrm dx$

or in terms of $\log_{1/e}x$,

$\displaystyle\int_1^e\ln x-\log_{1/e}x\,\mathrm dx$