What is the integral from 0 to (7pi)/6 of 2sin(x)+1 ?

The integral from 0 to (7pi)/6 of 2sin(x)+1 is sqrt(3)+2+(7pi)/6

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integral from 0 to (7pi)/6 of 2sin(x)+1

\int_{0}^{\frac{7 π}{6}} 2 sin (x) + 1 d x

3 + 2 + \frac{7 π}{6}

Decimal

7.39724 \dots

Solution steps

\int_{0}^{\frac{7 π}{6}} 2 sin (x) + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{\frac{7 π}{6}} 2 sin (x) d x + \int_{0}^{\frac{7 π}{6}} 1 d x

\int_{0}^{\frac{7 π}{6}} 2 sin (x) d x = 3 + 2

\int_{0}^{\frac{7 π}{6}} 1 d x = \frac{7 π}{6}

= 3 + 2 + \frac{7 π}{6}

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What is the integral from 0 to (7pi)/6 of 2sin(x)+1 ?
The integral from 0 to (7pi)/6 of 2sin(x)+1 is sqrt(3)+2+(7pi)/6