What is the integral from 0 to ln(4) of e^{3x} ?

The integral from 0 to ln(4) of e^{3x} is 21

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integral from 0 to ln(4) of e^{3x}

\int_{0}^{l n (4)} e^{3 x} d x

21

Solution steps

\int_{0}^{l n (4)} e^{3 x} d x

Apply u-substitution

: \int_{0}^{6 l n (2)} e^{u} \frac{1}{3} d u

= \int_{0}^{6 l n (2)} e^{u} \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{6 l n (2)} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{0}^{6 l n (2)}

Compute the boundaries:

63

= \frac{1}{3} \cdot 63

Simplify

= 21

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\int_{2}^{4} 7 x ln (x) d x

\int_{1}^{\infty} \frac{1}{7 ^{x}} d x

What is the integral from 0 to ln(4) of e^{3x} ?
The integral from 0 to ln(4) of e^{3x} is 21