What is the integral from 0 to 1 of (e^{2x}-1)/2 ?

The integral from 0 to 1 of (e^{2x}-1)/2 is (e^2-3)/4

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integral from 0 to 1 of (e^{2x}-1)/2

\int_{0}^{1} \frac{e ^{2 x} - 1}{2} d x

\frac{e ^{2} - 3}{4}

Decimal Notation

1.09726 \dots

Solution steps

\int_{0}^{1} \frac{e ^{2 x} - 1}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} e^{2 x} - 1 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

= \frac{1}{2} (\int_{0}^{1} e^{2 x} d x - \int_{0}^{1} 1 d x)

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

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

= \frac{1}{2} (\frac{e ^{2} - 1}{2} - 1)

Simplify

\frac{1}{2} (\frac{e ^{2} - 1}{2} - 1) : \frac{e ^{2} - 3}{4}

= \frac{e ^{2} - 3}{4}

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What is the integral from 0 to 1 of (e^{2x}-1)/2 ?
The integral from 0 to 1 of (e^{2x}-1)/2 is (e^2-3)/4