What is the integral from 2 to 5 of (6+cos(x/(10))) ?

The integral from 2 to 5 of (6+cos(x/(10))) is 18+10(sin(1/2)-sin(1/5))

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integral from 2 to 5 of (6+cos(x/(10)))

\int_{2}^{5} (6 + cos (\frac{x}{10})) d x

18 + 10 (sin (\frac{1}{2}) - sin (\frac{1}{5}))

Decimal

20.80756 \dots

Solution steps

\int_{2}^{5} 6 + cos (\frac{x}{10}) 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_{2}^{5} 6 d x + \int_{2}^{5} cos (\frac{x}{10}) d x

\int_{2}^{5} 6 d x = 18

\int_{2}^{5} cos (\frac{x}{10}) d x = 10 (sin (\frac{1}{2}) - sin (\frac{1}{5}))

= 18 + 10 (sin (\frac{1}{2}) - sin (\frac{1}{5}))

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What is the integral from 2 to 5 of (6+cos(x/(10))) ?
The integral from 2 to 5 of (6+cos(x/(10))) is 18+10(sin(1/2)-sin(1/5))