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

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

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

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

2

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

\int sin (x) d x = - cos (x)

= 2 [- cos (x)]_{0}^{\frac{3 π}{2}}

Compute the boundaries:

1

= 2 \cdot 1

Simplify

= 2

\int_{0}^{6} 2.8 x d x

\int_{0}^{π} (sin^{3} (θ)) d θ

\int_{2}^{5} cos (x) + e^{x} - 2 x d x

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

\int_{5}^{5} 1

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