How do you evaluate the integral of $\int | sin x | d x$ $int abssinx dx$ from 0 to 3pi/2?

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How do you evaluate the integral of $\int | sin x | d x$ $int abssinx dx$ from 0 to 3pi/2?

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Answer 1 · 2016-03-24T15:12:00

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Explanation:

Remark that
$| sin x | = sin x$ $|sinx|=sinx$ for $0 \leq x \leq π$ $0 <= x <=pi$
$| sin x | = - sin x$ $|sinx|=-sinx$ for $π \leq x \leq \frac{3 π}{2}$ $pi<=x<=(3pi)/2$

Therefore we can rewrite the expression as
$= \int_{0}^{π} sin x \cdot d x - \int_{π}^{\frac{3 π}{2}} sin x \cdot d x$ $=int_0^pi sinx*dx-int_pi^((3pi)/2) sinx*dx$
$= - cos x {∣ ∣}_{0}^{π}$ $=-cos x|_0^pi$ $+ cos x {∣ ∣ ∣}_{π}^{\frac{3 π}{2}}$ $+cosx|_pi^((3pi)/2)$
$= - [- 1 - (1)] + [0 - (- 1)] = 2 + 1 = 3$ $=-[-1-(1)]+[0-(-1)]=2+1=3$

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How do you evaluate the integral of $\int | sin x | d x$ $int abssinx dx$ from 0 to 3pi/2?

1 Answer

Explanation:

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How do you evaluate the integral of $\int | sin x | d x$ $int abssinx dx$ from 0 to 3pi/2?