What is the integral from 0 to ln(4) of e^{-2x} ?

The integral from 0 to ln(4) of e^{-2x} is 15/32

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integral from 0 to ln(4) of e^{-2x}

\int_{0}^{l n (4)} e^{- 2 x} d x

\frac{15}{32}

Decimal Notation

0.46875

Solution steps

\int_{0}^{l n (4)} e^{- 2 x} d x

Apply u-substitution

: \int_{0}^{- 4 l n (2)} - \frac{1}{2} e^{u} d u

= \int_{0}^{- 4 l n (2)} - \frac{1}{2} e^{u} d u

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

= - \int_{- 4 l n (2)}^{0} - \frac{1}{2} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{- 4 l n (2)}^{0} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{2} [e^{u}]_{- 4 l n (2)}^{0})

Simplify

= \frac{1}{2} [e^{u}]_{- 4 l n (2)}^{0}

Compute the boundaries:

\frac{15}{16}

= \frac{1}{2} \cdot \frac{15}{16}

Simplify

= \frac{15}{32}

\frac{d}{d x} (e^{t a n (x)})

\frac{d}{d x} (x 400 - x^{2})

5 y^{^{'}} + 15 y = 375, y (0) = 0

e^{x}, 1

n = 0 \sum \infty - \frac{n}{2 ^{n}}

What is the integral from 0 to ln(4) of e^{-2x} ?
The integral from 0 to ln(4) of e^{-2x} is 15/32