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

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

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

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

- \frac{15}{32}

Decimal

- 0.46875

Solution steps

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

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

= - \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}

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