What is the integral from 0 to 3 of e^{-1/2 t} ?

The integral from 0 to 3 of e^{-1/2 t} is 2(1-1/(e^{3/2)})

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integral from 0 to 3 of e^{-1/2 t}

\int_{0}^{3} e^{- \frac{1}{2} t} d t

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

Decimal

1.55373 \dots

Solution steps

\int_{0}^{3} e^{- \frac{1}{2} t} d t

Apply u-substitution

: \int_{0}^{- \frac{3}{2}} - 2 e^{u} d u

= \int_{0}^{- \frac{3}{2}} - 2 e^{u} d u

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

= - \int_{- \frac{3}{2}}^{0} - 2 e^{u} d u

Take the constant out:

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

= - (- 2 \cdot \int_{- \frac{3}{2}}^{0} e^{u} d u)

Use the common integral:

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

= - (- 2 [e^{u}]_{- \frac{3}{2}}^{0})

Simplify

= 2 [e^{u}]_{- \frac{3}{2}}^{0}

Compute the boundaries:

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

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

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What is the integral from 0 to 3 of e^{-1/2 t} ?
The integral from 0 to 3 of e^{-1/2 t} is 2(1-1/(e^{3/2)})