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

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

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

\int_{0}^{l n (2)} e^{2 x} - e^{x} d x

\frac{1}{2}

Decimal Notation

0.5

Solution steps

\int_{0}^{l n (2)} 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_{0}^{l n (2)} e^{2 x} d x - \int_{0}^{l n (2)} e^{x} d x

\int_{0}^{l n (2)} e^{2 x} d x = \frac{3}{2}

\int_{0}^{l n (2)} e^{x} d x = 1

= \frac{3}{2} - 1

Simplify

\frac{3}{2} - 1 : \frac{1}{2}

= \frac{1}{2}

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