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

The integral from-1 to 1 of 4/(3e^{3x)} is (4(e^6-1))/(9e^3)

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integral from-1 to 1 of 4/(3e^{3x)}

\int_{- 1}^{1} \frac{4}{3 e ^{3 x}} d x

\frac{4 ( e ^{6} - 1 )}{9 e ^{3}}

Decimal

8.90477 \dots

Solution steps

\int_{- 1}^{1} \frac{4}{3 e ^{3 x}} d x

Take the constant out:

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

= \frac{4}{3} \cdot \int_{- 1}^{1} \frac{1}{e ^{3 x}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= \frac{4}{3} \cdot \int_{- 1}^{1} e^{- 3 x} d x

Apply u-substitution

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

= \frac{4}{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

= \frac{4}{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

= \frac{4}{3} (- (- \frac{1}{3} \cdot \int_{- 3}^{3} e^{u} d u))

Use the common integral:

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

= \frac{4}{3} (- (- \frac{1}{3} [e^{u}]_{- 3}^{3}))

Simplify

\frac{4}{3} (- (- \frac{1}{3} [e^{u}]_{- 3}^{3})) : \frac{4}{9} [e^{u}]_{- 3}^{3}

= \frac{4}{9} [e^{u}]_{- 3}^{3}

Compute the boundaries:

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

= \frac{4}{9} (e^{3} - \frac{1}{e ^{3}})

Simplify

= \frac{4 ( e ^{6} - 1 )}{9 e ^{3}}

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What is the integral from-1 to 1 of 4/(3e^{3x)} ?
The integral from-1 to 1 of 4/(3e^{3x)} is (4(e^6-1))/(9e^3)