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

The integral from-1 to 1 of 2e^{-3t+8} is (2(e^{11}-e^5))/3

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

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

\frac{2 ( e ^{11} - e ^{5} )}{3}

Decimal

39817.15237 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{11}^{5} - \frac{1}{3} e^{u} d u

= 2 \cdot \int_{11}^{5} - \frac{1}{3} e^{u} d u

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

= 2 (- \int_{5}^{11} - \frac{1}{3} e^{u} d u)

Take the constant out:

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

= 2 (- (- \frac{1}{3} \cdot \int_{5}^{11} e^{u} d u))

Use the common integral:

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

= 2 (- (- \frac{1}{3} [e^{u}]_{5}^{11}))

Simplify

2 (- (- \frac{1}{3} [e^{u}]_{5}^{11})) : \frac{2}{3} [e^{u}]_{5}^{11}

= \frac{2}{3} [e^{u}]_{5}^{11}

Compute the boundaries:

e^{11} - e^{5}

= \frac{2}{3} (e^{11} - e^{5})

Simplify

= \frac{2 ( e ^{11} - e ^{5} )}{3}

\int_{0}^{1} e^{- 5 x} d x

\int_{0}^{1} x^{7} e^{x^{4}} d x

\int_{1}^{2} [\frac{1}{x} - \frac{2}{x ^{3}}] d x

\int_{0}^{\frac{π}{2}} 0 d x

\int_{1}^{3} - \frac{x}{3} + 1 d x

What is the integral from-1 to 1 of 2e^{-3t+8} ?
The integral from-1 to 1 of 2e^{-3t+8} is (2(e^{11}-e^5))/3