What is the integral from 0 to ln(2) of e^{-3x} ?

The integral from 0 to ln(2) of e^{-3x} is 7/24

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integral from 0 to ln(2) of e^{-3x}

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

\frac{7}{24}

Decimal

0.29166 \dots

Solution steps

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

Apply u-substitution

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

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

\frac{7}{8}

= \frac{1}{3} \cdot \frac{7}{8}

Simplify

= \frac{7}{24}

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What is the integral from 0 to ln(2) of e^{-3x} ?
The integral from 0 to ln(2) of e^{-3x} is 7/24