What is the integral from 2 to 3 of-8e^{-2x+1} ?

The integral from 2 to 3 of-8e^{-2x+1} is -4(1/(e^3)-1/(e^5))

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integral from 2 to 3 of-8e^{-2x+1}

\int_{2}^{3} - 8 e^{- 2 x + 1} d x

- 4 (\frac{1}{e ^{3}} - \frac{1}{e ^{5}})

Decimal

- 0.17219 \dots

Solution steps

\int_{2}^{3} - 8 e^{- 2 x + 1} d x

Take the constant out:

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

= - 8 \cdot \int_{2}^{3} e^{- 2 x + 1} d x

Apply u-substitution

: \int_{- 3}^{- 5} - \frac{1}{2} e^{u} d u

= - 8 \cdot \int_{- 3}^{- 5} - \frac{1}{2} e^{u} d u

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

= - 8 (- \int_{- 5}^{- 3} - \frac{1}{2} e^{u} d u)

Take the constant out:

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

= - 8 (- (- \frac{1}{2} \cdot \int_{- 5}^{- 3} e^{u} d u))

Use the common integral:

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

= - 8 (- (- \frac{1}{2} [e^{u}]_{- 5}^{- 3}))

Simplify

- 8 (- (- \frac{1}{2} [e^{u}]_{- 5}^{- 3})) : - 4 [e^{u}]_{- 5}^{- 3}

= - 4 [e^{u}]_{- 5}^{- 3}

Compute the boundaries:

\frac{1}{e ^{3}} - \frac{1}{e ^{5}}

= - 4 (\frac{1}{e ^{3}} - \frac{1}{e ^{5}})

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What is the integral from 2 to 3 of-8e^{-2x+1} ?
The integral from 2 to 3 of-8e^{-2x+1} is -4(1/(e^3)-1/(e^5))