What is the integral from 0 to 1 of (e^x+e^{3x}) ?

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

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integral from 0 to 1 of (e^x+e^{3x})

\int_{0}^{1} (e^{x} + e^{3 x}) d x

e - 1 + \frac{e ^{3} - 1}{3}

Decimal Notation

8.08012 \dots

Solution steps

\int_{0}^{1} e^{x} + e^{3 x} 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

= \int_{0}^{1} e^{x} d x + \int_{0}^{1} e^{3 x} d x

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

\int_{0}^{1} e^{3 x} d x = \frac{e ^{3} - 1}{3}

= e - 1 + \frac{e ^{3} - 1}{3}

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