What is the integral from pi/2 to pi of 1/2 e^{-θ/4} ?

The integral from pi/2 to pi of 1/2 e^{-θ/4} is 2(1/(e^{pi/8)}-1/(e^{pi/4)})

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integral from pi/2 to pi of 1/2 e^{-θ/4}

\int_{\frac{π}{2}}^{π} \frac{1}{2} e^{- \frac{θ}{4}} d θ

2 (\frac{1}{e ^{\frac{π}{8}}} - \frac{1}{e ^{\frac{π}{4}}})

Decimal Notation

0.43858 \dots

Solution steps

\int_{\frac{π}{2}}^{π} \frac{1}{2} e^{- \frac{θ}{4}} d θ

Take the constant out:

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

= \frac{1}{2} \cdot \int_{\frac{π}{2}}^{π} e^{- \frac{θ}{4}} d θ

Apply u-substitution

: \int_{- \frac{π}{8}}^{- \frac{π}{4}} - 4 e^{u} d u

= \frac{1}{2} \cdot \int_{- \frac{π}{8}}^{- \frac{π}{4}} - 4 e^{u} d u

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

= \frac{1}{2} (- \int_{- \frac{π}{4}}^{- \frac{π}{8}} - 4 e^{u} d u)

Take the constant out:

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

= \frac{1}{2} (- (- 4 \cdot \int_{- \frac{π}{4}}^{- \frac{π}{8}} e^{u} d u))

Use the common integral:

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

= \frac{1}{2} (- (- 4 [e^{u}]_{- \frac{π}{4}}^{- \frac{π}{8}}))

Simplify

\frac{1}{2} (- (- 4 [e^{u}]_{- \frac{π}{4}}^{- \frac{π}{8}})) : 2 [e^{u}]_{- \frac{π}{4}}^{- \frac{π}{8}}

= 2 [e^{u}]_{- \frac{π}{4}}^{- \frac{π}{8}}

Compute the boundaries:

\frac{1}{e ^{\frac{π}{8}}} - \frac{1}{e ^{\frac{π}{4}}}

= 2 (\frac{1}{e ^{\frac{π}{8}}} - \frac{1}{e ^{\frac{π}{4}}})

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What is the integral from pi/2 to pi of 1/2 e^{-θ/4} ?
The integral from pi/2 to pi of 1/2 e^{-θ/4} is 2(1/(e^{pi/8)}-1/(e^{pi/4)})