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

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

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

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

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

Decimal Notation

0.34956 \dots

Solution steps

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

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

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

\int_{- 1}^{2} - x^{2} d x

\int_{- 2}^{2} - x^{2} + 5 d x

\int_{0}^{1} \frac{1}{2} e^{x} + \frac{3}{4} 2^{x} d x

\int_{- 3}^{7} (∣ x ∣) d x

\int_{0}^{\frac{π}{2}} sin (8 x) d x

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