What is the integral from 0 to (3pi)/2 of 9|sin(x)| ?

The integral from 0 to (3pi)/2 of 9|sin(x)| is 27

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integral from 0 to (3pi)/2 of 9|sin(x)|

\int_{0}^{\frac{3 π}{2}} 9 ∣ sin (x) ∣ d x

27

Solution steps

\int_{0}^{\frac{3 π}{2}} 9 ∣ sin (x) ∣ d x

Eliminate Absolutes

= \int_{0}^{π} 9 sin (x) d x + \int_{π}^{\frac{3 π}{2}} 9 (- sin (x)) d x

\int_{0}^{π} 9 sin (x) d x = 18

\int_{π}^{\frac{3 π}{2}} 9 (- sin (x)) d x = 9

= 18 + 9

Simplify

= 27

\int \frac{e ^{2 x}}{- 2} d x

\int_{0}^{\infty} \frac{1}{9 + x ^{2}} d x

\int \frac{1}{( x ^{2} + 2 x )} d x

x^{2} - x^{3} + 4

x \to \infty lim (x^{3} - 1)

What is the integral from 0 to (3pi)/2 of 9|sin(x)| ?
The integral from 0 to (3pi)/2 of 9|sin(x)| is 27