What is the integral from-2 to 3 of e^{(-4x)/3} ?

The integral from-2 to 3 of e^{(-4x)/3} is (3(e^{20/3}-1))/(4e^4)

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integral from-2 to 3 of e^{(-4x)/3}

\int_{- 2}^{3} e^{\frac{- 4 x}{3}} d x

\frac{3 ( e ^{\frac{20}{3}} - 1 )}{4 e ^{4}}

Decimal

10.78020 \dots

Solution steps

\int_{- 2}^{3} e^{\frac{- 4 x}{3}} d x

Simplify

= \int_{- 2}^{3} e^{- \frac{4 x}{3}} d x

Apply u-substitution

: \int_{\frac{8}{3}}^{- 4} - \frac{3}{4} e^{u} d u

= \int_{\frac{8}{3}}^{- 4} - \frac{3}{4} e^{u} d u

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

= - \int_{- 4}^{\frac{8}{3}} - \frac{3}{4} e^{u} d u

Take the constant out:

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

= - (- \frac{3}{4} \cdot \int_{- 4}^{\frac{8}{3}} e^{u} d u)

Use the common integral:

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

= - (- \frac{3}{4} [e^{u}]_{- 4}^{\frac{8}{3}})

Simplify

= \frac{3}{4} [e^{u}]_{- 4}^{\frac{8}{3}}

Compute the boundaries:

e^{\frac{8}{3}} - \frac{1}{e ^{4}}

= \frac{3}{4} (e^{\frac{8}{3}} - \frac{1}{e ^{4}})

Simplify

= \frac{3 ( e ^{\frac{20}{3}} - 1 )}{4 e ^{4}}

\int_{1}^{4} (\frac{x}{8} - \frac{3}{x}) d x

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\int_{1}^{3} h (x) d x

\int_{1}^{3} x e^{2 x} d x

\int_{4}^{x} \frac{1}{1 + t + t ^{2}} d t

What is the integral from-2 to 3 of e^{(-4x)/3} ?
The integral from-2 to 3 of e^{(-4x)/3} is (3(e^{20/3}-1))/(4e^4)