What is the integral from-2 to 0 of-5e^{-7x+5} ?

The integral from-2 to 0 of-5e^{-7x+5} is -(5(e^{19}-e^5))/7

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integral from-2 to 0 of-5e^{-7x+5}

\int_{- 2}^{0} - 5 e^{- 7 x + 5} d x

- \frac{5 ( e ^{19} - e ^{5} )}{7}

Decimal

- 127487251.82144 \dots

Solution steps

\int_{- 2}^{0} - 5 e^{- 7 x + 5} d x

Take the constant out:

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

= - 5 \cdot \int_{- 2}^{0} e^{- 7 x + 5} d x

Apply u-substitution

: \int_{19}^{5} - \frac{1}{7} e^{u} d u

= - 5 \cdot \int_{19}^{5} - \frac{1}{7} e^{u} d u

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

= - 5 (- \int_{5}^{19} - \frac{1}{7} e^{u} d u)

Take the constant out:

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

= - 5 (- (- \frac{1}{7} \cdot \int_{5}^{19} e^{u} d u))

Use the common integral:

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

= - 5 (- (- \frac{1}{7} [e^{u}]_{5}^{19}))

Simplify

- 5 (- (- \frac{1}{7} [e^{u}]_{5}^{19})) : - \frac{5}{7} [e^{u}]_{5}^{19}

= - \frac{5}{7} [e^{u}]_{5}^{19}

Compute the boundaries:

e^{19} - e^{5}

= - \frac{5}{7} (e^{19} - e^{5})

Simplify

= - \frac{5 ( e ^{19} - e ^{5} )}{7}

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What is the integral from-2 to 0 of-5e^{-7x+5} ?
The integral from-2 to 0 of-5e^{-7x+5} is -(5(e^{19}-e^5))/7