What is the integral from pi/2 to pi of 1/2 e^{-θ/6} ?

The integral from pi/2 to pi of 1/2 e^{-θ/6} is 3(1/(e^{\frac{pi){12}}}-1/(e^{pi/6)})

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integral from pi/2 to pi of 1/2 e^{-θ/6}

\int_{\frac{π}{2}}^{π} \frac{1}{2} e^{- \frac{θ}{6}} d θ

3 (\frac{1}{e ^{\frac{π}{12}}} - \frac{1}{e ^{\frac{π}{6}}})

Decimal

0.53184 \dots

Solution steps

\int_{\frac{π}{2}}^{π} \frac{1}{2} e^{- \frac{θ}{6}} 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}{6}} d θ

Apply u-substitution

: \int_{- \frac{π}{12}}^{- \frac{π}{6}} - 6 e^{u} d u

= \frac{1}{2} \cdot \int_{- \frac{π}{12}}^{- \frac{π}{6}} - 6 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}{6}}^{- \frac{π}{12}} - 6 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} (- (- 6 \cdot \int_{- \frac{π}{6}}^{- \frac{π}{12}} e^{u} d u))

Use the common integral:

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

= \frac{1}{2} (- (- 6 [e^{u}]_{- \frac{π}{6}}^{- \frac{π}{12}}))

Simplify

\frac{1}{2} (- (- 6 [e^{u}]_{- \frac{π}{6}}^{- \frac{π}{12}})) : 3 [e^{u}]_{- \frac{π}{6}}^{- \frac{π}{12}}

= 3 [e^{u}]_{- \frac{π}{6}}^{- \frac{π}{12}}

Compute the boundaries:

\frac{1}{e ^{\frac{π}{12}}} - \frac{1}{e ^{\frac{π}{6}}}

= 3 (\frac{1}{e ^{\frac{π}{12}}} - \frac{1}{e ^{\frac{π}{6}}})

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What is the integral from pi/2 to pi of 1/2 e^{-θ/6} ?
The integral from pi/2 to pi of 1/2 e^{-θ/6} is 3(1/(e^{\frac{pi){12}}}-1/(e^{pi/6)})