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

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

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

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

\frac{e - 1}{2}

Decimal Notation

0.85914 \dots

Solution steps

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

Apply exponent rule:

a^{n} \cdot b^{n} = (a \cdot b)^{n}

e^{x} e^{x} = (ee)^{x}

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

\int a^{x} d x = \frac{a ^{x}}{ln a}

= [\frac{( ee ) ^{x}}{ln ( ee )}]_{0}^{\frac{1}{2}}

Simplify

[\frac{( ee ) ^{x}}{ln ( ee )}]_{0}^{\frac{1}{2}} : [\frac{1}{2} e^{2 x}]_{0}^{\frac{1}{2}}

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

Compute the boundaries:

\frac{e}{2} - \frac{1}{2}

= \frac{e}{2} - \frac{1}{2}

Simplify

= \frac{e - 1}{2}

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