What is the integral from-pi to pi of |sin(t)| ?

The integral from-pi to pi of |sin(t)| is 4

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integral from-pi to pi of |sin(t)|

\int_{- π}^{π} ∣ sin (t) ∣ d t

4

Solution steps

\int_{- π}^{π} ∣ sin (t) ∣ d t

Eliminate Absolutes

= \int_{- π}^{0} - sin (t) d t + \int_{0}^{π} sin (t) d t

\int_{- π}^{0} - sin (t) d t = 2

\int_{0}^{π} sin (t) d t = 2

= 2 + 2

Simplify

= 4

\int_{0}^{1} x^{\frac{1}{2}} - x^{2} d x

\int_{0}^{x} \frac{3 + t ^{2}}{1 - t ^{4}} d t

\int_{1}^{2} \frac{x ^{2} - a}{x} d x

\int_{- 1}^{1} (x - 3 + \frac{1}{x - 2}) d x

\int_{- 3}^{2} (9 - x^{2}) - (x + 3) d x

What is the integral from-pi to pi of |sin(t)| ?
The integral from-pi to pi of |sin(t)| is 4