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

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

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

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

\frac{π}{2} + 1

Decimal

2.57079 \dots

Solution steps

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

\int_{0}^{\frac{π}{2}} 1 d x = \frac{π}{2}

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

= \frac{π}{2} + 1

\int_{0}^{1} sin (t) d t

\int_{1}^{3} (2 - x)^{2} d x

\int_{0}^{4} (\frac{1}{2 x + 1}) d x

\int_{2 π}^{3 π} sin (x) d x

\int_{0}^{2} e^{- y} d y

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