What is the integral from-1 to 0 of 3e^{-3x} ?

The integral from-1 to 0 of 3e^{-3x} is e^3-1

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integral from-1 to 0 of 3e^{-3x}

\int_{- 1}^{0} 3 e^{- 3 x} d x

e^{3} - 1

Decimal Notation

19.08553 \dots

Solution steps

\int_{- 1}^{0} 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}^{0} e^{- 3 x} d x

Apply u-substitution

: \int_{3}^{0} - \frac{1}{3} e^{u} d u

= 3 \cdot \int_{3}^{0} - \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_{0}^{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_{0}^{3} e^{u} d u))

Use the common integral:

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

= 3 (- (- \frac{1}{3} [e^{u}]_{0}^{3}))

Simplify

3 (- (- \frac{1}{3} [e^{u}]_{0}^{3})) : [e^{u}]_{0}^{3}

= [e^{u}]_{0}^{3}

Compute the boundaries:

e^{3} - 1

= e^{3} - 1

f (x) = \frac{1 + 2 x}{1 - 2 x} e n e l, (1, - 3)

x \to 2 lim (\frac{2}{( x + 2 ) ^{2}})

\frac{e ^{x}}{x}

\int tan (x) \cdot sec^{2} (x) d x

\frac{s}{( s - 1 ) ( s ^{2} + 2 s + 2 )}

What is the integral from-1 to 0 of 3e^{-3x} ?
The integral from-1 to 0 of 3e^{-3x} is e^3-1