What is the integral from 0 to ln(4) of e^{6x} ?

The integral from 0 to ln(4) of e^{6x} is 1365/2

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integral from 0 to ln(4) of e^{6x}

\int_{0}^{l n (4)} e^{6 x} d x

\frac{1365}{2}

Decimal

682.5

Solution steps

\int_{0}^{l n (4)} e^{6 x} d x

Apply u-substitution

: \int_{0}^{12 l n (2)} e^{u} \frac{1}{6} d u

= \int_{0}^{12 l n (2)} e^{u} \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{0}^{12 l n (2)} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{6} [e^{u}]_{0}^{12 l n (2)}

Compute the boundaries:

4095

= \frac{1}{6} \cdot 4095

Simplify

= \frac{1365}{2}

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What is the integral from 0 to ln(4) of e^{6x} ?
The integral from 0 to ln(4) of e^{6x} is 1365/2