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

The integral from 0 to 2pi of 2x is 4pi^2

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integral from 0 to 2pi of 2x

\int_{0}^{2 π} 2 x d x

4 π^{2}

Decimal

39.47841 \dots

Solution steps

\int_{0}^{2 π} 2 x d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{2 π} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{2 π}

= 2 [\frac{x ^{2}}{2}]_{0}^{2 π}

Compute the boundaries:

2 π^{2}

= 2 \cdot 2 π^{2}

Simplify

= 4 π^{2}

y = x (x^{3} + 7)^{4}

\frac{\partial}{\partial x} (\frac{1}{2 x + 1})

n = 0 \sum \infty (- \frac{3}{5})^{n}

\int (9 + x^{2})^{- 2} d x

f (x) = x^{3} - 3 x + 1

What is the integral from 0 to 2pi of 2x ?
The integral from 0 to 2pi of 2x is 4pi^2