What is the integral from 0 to 3 of 3e^{-2x} ?

The integral from 0 to 3 of 3e^{-2x} is (3(e^6-1))/(2e^6)

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integral from 0 to 3 of 3e^{-2x}

\int_{0}^{3} 3 e^{- 2 x} d x

\frac{3 ( e ^{6} - 1 )}{2 e ^{6}}

Decimal Notation

1.49628 \dots

Solution steps

\int_{0}^{3} 3 e^{- 2 x} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{3} e^{- 2 x} d x

Apply u-substitution

: \int_{0}^{- 6} - \frac{1}{2} e^{u} d u

= 3 \cdot \int_{0}^{- 6} - \frac{1}{2} e^{u} d u

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

= 3 (- \int_{- 6}^{0} - \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

= 3 (- (- \frac{1}{2} \cdot \int_{- 6}^{0} e^{u} d u))

Use the common integral:

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

= 3 (- (- \frac{1}{2} [e^{u}]_{- 6}^{0}))

Simplify

3 (- (- \frac{1}{2} [e^{u}]_{- 6}^{0})) : \frac{3}{2} [e^{u}]_{- 6}^{0}

= \frac{3}{2} [e^{u}]_{- 6}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{6}}

= \frac{3}{2} (1 - \frac{1}{e ^{6}})

Simplify

= \frac{3 ( e ^{6} - 1 )}{2 e ^{6}}

sin^{3} (t)

\int x 5^{- x^{2}} d x

sec (x)

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

(6, - 2), (12, 1)

What is the integral from 0 to 3 of 3e^{-2x} ?
The integral from 0 to 3 of 3e^{-2x} is (3(e^6-1))/(2e^6)