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

The integral from-1 to 1 of 3e^{-3x} is e^3-1/(e^3)

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

\int_{- 1}^{1} 3 e^{- 3 x} d x

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

Decimal Notation

20.03574 \dots

Solution steps

\int_{- 1}^{1} 3 e^{- 3 x} d x

Take the constant out:

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

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

Apply u-substitution

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

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

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

= 3 (- \int_{- 3}^{3} - \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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

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

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

5 e^{- 2 t}

y^{^{'}} = 12 t^{5} - 20 t^{4} + 2 - 6 t^{- 2}

y = t e^{- 3 t} sin (5 t)

\frac{( 3 x - 7 )}{( 9 x + 5 )} + x^{3}

\frac{\partial}{\partial y} (x^{2} y)

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