If a definite integral represents the area under the curve, why is $\int_{0}^{\pi/2} \ln|\sin x| \, dx = -\pi \ln(2)/2$?

Question

I just noticed that $\int_{0}^{\pi/2} \ln|\sin x| \, dx = -\pi \ln(2)/2$ and my question is why is the area in this case a negative number?

It's the signed area, the parts below the y-axis count with a negative sign. — Florian, Jun 28 '19 at 07:21
A definite integral does not represent the area under the curve, but rather the signed area under the curve. — Matti P., Jun 28 '19 at 07:21

score 1 · Answer 1 · answered Jun 28 '19 at 07:22

1

The definite integral $\int_a^{b}f(x)dx$ represents the area under the graph of $f$ is $f$ is a positive function. Here the integrand is negative and the integral is the negative of the area between the graph and the $x-$ axis.

answered Jun 28 '19 at 07:22

Kavi Rama Murthy

311,013

score 0 · Accepted Answer · edited Jun 28 '19 at 08:38

0

If $f:[a,b] \to \mathbb R$ is Riemann-integrable and if $f \le 0$ on $[a,b]$, then the area between the curve and the x - axis is given by

$$\left|\int_a^b f(x) \mathrm{d}x\right|.$$

edited Jun 28 '19 at 08:38

Peter Foreman

19,947

answered Jun 28 '19 at 07:23

Fred

77,394

If a definite integral represents the area under the curve, why is $\int_{0}^{\pi/2} \ln|\sin x| \, dx = -\pi \ln(2)/2$?

2 Answers2