Definition of natural logarithm functions

Question

My textbook, University Calculus by Hass, Heil, Bogacki, Weir and Thomas, defines the natural log of x by its domain. It states that “from the fundamental theorem of calculus, we know that ln(x) is a continuous function. Geometrically, if x>1, then ln(x) is the are under the curve y=(1/t) from t=1 to t=x. For 0<x<1, ln(x) gives the negative of the area under the curve from x to 1.” My question is: what do the writers mean by the negative of the area under the curve, if the curve always remains above the x axis. Thanks

Answer 1

Welcome. You’re right the curve remains above the $x$-axis so its area is, unambiguously, a positive quantity. However the logarithm is negative on the interval $0<x<1$: you take the area and multiply it by $(-1)$. They did not mean you take the negative area in the sense you think, but rather you take the negative of the area.

Answer 2

If for $0<x<1$ you calculate $$\int_x^1 \frac1w \, dw=\log_e(1)-\log_e(x)=-\log_e(x)$$ then this is the area under the curve from $x$ to $1$ and is positive since $\log_e(x) <0$ for this $x$.

They are saying that reversing the limits $$\int_1^x \frac1w \, dw=\log_e(x)-\log_e(1)=\log_e(x)$$ even if $0<x<1$ and this is then the negative of the area under the curve from $x$ to $1$.