What is the integral from 0 to 1 of (2x) ?

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

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integral from 0 to 1 of (2x)

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

1

Solution steps

\int_{0}^{1} 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}^{1} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{1}

= 2 [\frac{x ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= 2 \cdot \frac{1}{2}

Simplify

= 1

\frac{4 x - 9}{x - 9}

x \to 0 - lim (x^{2} + \frac{1}{x})

\frac{2}{1 - x ^{2}}

\int \frac{1}{4 e ^{- 6 x} + e ^{6 x}} d x

g (t) = \frac{3}{t ^{2}}

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