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

The integral from ln(2) to ln(4) of e^{-x} is 1/4

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

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

\frac{1}{4}

Decimal

0.25

Solution steps

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

Apply u-substitution

: \int_{- l n (2)}^{- 2 l n (2)} - e^{u} d u

= \int_{- l n (2)}^{- 2 l n (2)} - e^{u} d u

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

= - \int_{- 2 l n (2)}^{- l n (2)} - e^{u} d u

Take the constant out:

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

= - (- \int_{- 2 l n (2)}^{- l n (2)} e^{u} d u)

Use the common integral:

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

= - (- [e^{u}]_{- 2 l n (2)}^{- l n (2)})

Simplify

= [e^{u}]_{- 2 l n (2)}^{- l n (2)}

Compute the boundaries:

\frac{1}{4}

= \frac{1}{4}

\frac{d}{d x} (\frac{e ^{x}}{sin ( x )})

h (x) = lo g_{10} (2 x^{2} + 5 x + 2)

(7 + x) y^{^{'}} + y = 7 + x

sec^{2} (x) - tan^{2} (x)

\frac{\partial}{\partial x} (\frac{x ^{4}}{6 y} + \frac{12 y ^{4}}{x})

What is the integral from ln(2) to ln(4) of e^{-x} ?
The integral from ln(2) to ln(4) of e^{-x} is 1/4