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=(\ln x)/(\ln\frac1e)=(\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<0`, 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`