What is the integral from 2 to 3 of (e^{2x}+e^x) ?

The integral from 2 to 3 of (e^{2x}+e^x) is (e^6-e^4)/2+e^3-e^2

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integral from 2 to 3 of (e^{2x}+e^x)

\int_{2}^{3} (e^{2 x} + e^{x}) d x

\frac{e ^{6} - e ^{4}}{2} + e^{3} - e^{2}

Decimal

187.11180 \dots

Solution steps

\int_{2}^{3} e^{2 x} + e^{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_{2}^{3} e^{2 x} d x + \int_{2}^{3} e^{x} d x

\int_{2}^{3} e^{2 x} d x = \frac{e ^{6} - e ^{4}}{2}

\int_{2}^{3} e^{x} d x = e^{3} - e^{2}

= \frac{e ^{6} - e ^{4}}{2} + e^{3} - e^{2}

What is the integral from 2 to 3 of (e^{2x}+e^x) ?
The integral from 2 to 3 of (e^{2x}+e^x) is (e^6-e^4)/2+e^3-e^2