What is the integral from 0 to 2pi of integral of 1 ?

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

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integral from 0 to 2pi of integral of 1

\int_{0}^{2 π} \int 1 d x d y

2 π (x + C)

Solution steps

\int_{0}^{2 π} \int 1 d x d y

\int 1 d x = x + C

= \int_{0}^{2 π} (x + C) d y

\int_{0}^{2 π} (x + C) d y = 2 π (x + C)

= 2 π (x + C)

\frac{\partial}{\partial z} (z^{3})

d er i v a t i v e sin (cos (x))

\int \frac{1}{3 x ( 4 + x )} d x

\int - 3 x^{2} e^{x} d x

\frac{d}{d x} (7 (3 x + \frac{x ^{\frac{1}{3}}}{3}))

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