What is the integral from ln(5) to ln(7) of e^{-x} ?

The integral from ln(5) to ln(7) of e^{-x} is 2/35

integral from ln(5) to ln(7) of e^{-x}

\int_{l n (5)}^{l n (7)} e^{- x} d x

\frac{2}{35}

Decimal

0.05714 \dots

Solution steps

\int_{l n (5)}^{l n (7)} e^{- x} d x

Apply u-substitution

= \int_{- l n (5)}^{- l n (7)} - e^{u} d u

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

= - \int_{- l n (7)}^{- l n (5)} - e^{u} d u

Take the constant out:

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

= - (- \int_{- l n (7)}^{- l n (5)} e^{u} d u)

Use the common integral:

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

= - (- [e^{u}]_{- l n (7)}^{- l n (5)})

Simplify

= [e^{u}]_{- l n (7)}^{- l n (5)}

Compute the boundaries:

\frac{2}{35}

= \frac{2}{35}

\int_{0}^{c} x (c - x) d x

\int_{- \frac{π}{2}}^{\frac{π}{2}} x d x

\int_{- 1}^{1} [(2 - x^{2}) - (x^{2})] d x

\int_{\frac{π}{3}}^{\frac{π}{2}} cos (π) x d x

\int_{0}^{π} (sin (t) cos (t)) d t

What is the integral from ln(5) to ln(7) of e^{-x} ?
The integral from ln(5) to ln(7) of e^{-x} is 2/35