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

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

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

\int_{0}^{π} e^{- \frac{t}{2}} d t

2 (1 - \frac{1}{e ^{\frac{π}{2}}})

Solution steps

\int_{0}^{π} e^{- \frac{t}{2}} d t

Apply u-substitution

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

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

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

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

Use the common integral:

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

= - (- 2 [e^{u}]_{- \frac{π}{2}}^{0})

Simplify

= 2 [e^{u}]_{- \frac{π}{2}}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{\frac{π}{2}}}

= 2 (1 - \frac{1}{e ^{\frac{π}{2}}})

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