What is the integral from 0 to 2pi of (1+sin(u)) ?

The integral from 0 to 2pi of (1+sin(u)) is 2pi

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integral from 0 to 2pi of (1+sin(u))

\int_{0}^{2 π} (1 + sin (u)) d u

2 π

Decimal

6.28318 \dots

Solution steps

\int_{0}^{2 π} 1 + sin (u) d u

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}^{2 π} 1 d u + \int_{0}^{2 π} sin (u) d u

\int_{0}^{2 π} 1 d u = 2 π

\int_{0}^{2 π} sin (u) d u = 0

= 2 π + 0

Simplify

= 2 π

y^{^{''}} = 5 y

\int_{a}^{b} 1 d x

\frac{d r}{d t} = - 2 t r

\frac{6}{s ^{2} + 4}

\int_{2}^{4} (2 - x^{2}) - (10 - 6 x) d x

What is the integral from 0 to 2pi of (1+sin(u)) ?
The integral from 0 to 2pi of (1+sin(u)) is 2pi