What is the integral from 0 to 4 of 8e^{-3x} ?

The integral from 0 to 4 of 8e^{-3x} is (8(e^{12}-1))/(3e^{12)}

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integral from 0 to 4 of 8e^{-3x}

\int_{0}^{4} 8 e^{- 3 x} d x

\frac{8 ( e ^{12} - 1 )}{3 e ^{12}}

Decimal Notation

2.66665 \dots

Solution steps

\int_{0}^{4} 8 e^{- 3 x} d x

Take the constant out:

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

= 8 \cdot \int_{0}^{4} e^{- 3 x} d x

Apply u-substitution

: \int_{0}^{- 12} - \frac{1}{3} e^{u} d u

= 8 \cdot \int_{0}^{- 12} - \frac{1}{3} e^{u} d u

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

= 8 (- \int_{- 12}^{0} - \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

= 8 (- (- \frac{1}{3} \cdot \int_{- 12}^{0} e^{u} d u))

Use the common integral:

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

= 8 (- (- \frac{1}{3} [e^{u}]_{- 12}^{0}))

Simplify

8 (- (- \frac{1}{3} [e^{u}]_{- 12}^{0})) : \frac{8}{3} [e^{u}]_{- 12}^{0}

= \frac{8}{3} [e^{u}]_{- 12}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{12}}

= \frac{8}{3} (1 - \frac{1}{e ^{12}})

Simplify

= \frac{8 ( e ^{12} - 1 )}{3 e ^{12}}

f (x) = x^{3} - x + 1, at x = 3

\int_{1}^{7} \frac{2}{x ^{3}} d x

tan (\frac{x}{2})

sin (3 x)

\int (cos^{4} (x)) d x

What is the integral from 0 to 4 of 8e^{-3x} ?
The integral from 0 to 4 of 8e^{-3x} is (8(e^{12}-1))/(3e^{12)}