What is the integral from 0 to 1 of e^{-4x} ?

The integral from 0 to 1 of e^{-4x} is (e^4-1)/(4e^4)

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integral from 0 to 1 of e^{-4x}

\int_{0}^{1} e^{- 4 x} d x

\frac{e ^{4} - 1}{4 e ^{4}}

Decimal Notation

0.24542 \dots

Solution steps

\int_{0}^{1} e^{- 4 x} d x

Apply u-substitution

: \int_{0}^{- 4} - \frac{1}{4} e^{u} d u

= \int_{0}^{- 4} - \frac{1}{4} e^{u} d u

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

= - \int_{- 4}^{0} - \frac{1}{4} e^{u} d u

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

= \frac{1}{4} [e^{u}]_{- 4}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{4}}

= \frac{1}{4} (1 - \frac{1}{e ^{4}})

Simplify

= \frac{e ^{4} - 1}{4 e ^{4}}

\frac{d}{d x} (C e^{- 2 x})

\int 3^{s i n (x)} cos (x) d x

P^{^{'}} = k P

9 t

\frac{d}{d x} (16 x^{\frac{3}{2}})

What is the integral from 0 to 1 of e^{-4x} ?
The integral from 0 to 1 of e^{-4x} is (e^4-1)/(4e^4)