What is the integral from-2 to 3 of e^{-t} ?

The integral from-2 to 3 of e^{-t} is e^2-1/(e^3)

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integral from-2 to 3 of e^{-t}

\int_{- 2}^{3} e^{- t} d t

e^{2} - \frac{1}{e ^{3}}

Decimal

7.33926 \dots

Solution steps

\int_{- 2}^{3} e^{- t} d t

Apply u-substitution

: \int_{2}^{- 3} - e^{u} d u

= \int_{2}^{- 3} - e^{u} d u

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

= - \int_{- 3}^{2} - e^{u} d u

Take the constant out:

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

= - (- \int_{- 3}^{2} e^{u} d u)

Use the common integral:

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

= - (- [e^{u}]_{- 3}^{2})

Simplify

= [e^{u}]_{- 3}^{2}

Compute the boundaries:

e^{2} - \frac{1}{e ^{3}}

= e^{2} - \frac{1}{e ^{3}}

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What is the integral from-2 to 3 of e^{-t} ?
The integral from-2 to 3 of e^{-t} is e^2-1/(e^3)