What is the integral from 0 to ln(5) of e^x(3-4e^x) ?

The integral from 0 to ln(5) of e^x(3-4e^x) is -36

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integral from 0 to ln(5) of e^x(3-4e^x)

\int_{0}^{l n (5)} e^{x} (3 - 4 e^{x}) d x

- 36

Solution steps

\int_{0}^{l n (5)} e^{x} (3 - 4 e^{x}) d x

Apply u-substitution

: \int_{- 1}^{- 17} - \frac{u}{4} d u

= \int_{- 1}^{- 17} - \frac{u}{4} d u

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

= - \int_{- 17}^{- 1} - \frac{u}{4} d u

Take the constant out:

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

= - (- \frac{1}{4} \cdot \int_{- 17}^{- 1} u d u)

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{- 17}^{- 1}

= - (- \frac{1}{4} [\frac{u ^{2}}{2}]_{- 17}^{- 1})

Simplify

= \frac{1}{4} [\frac{u ^{2}}{2}]_{- 17}^{- 1}

Compute the boundaries:

- 144

= \frac{1}{4} (- 144)

Simplify

= - 36

y = e^{2}, x = 0, x = 3

\frac{d}{d x} (x e^{r (x) x})

\frac{d}{d x} (lo g_{5} (arccos (x^{3} - 2)))

\frac{\partial}{\partial x} (- 2 e^{y} sin (- x) + 3 y - 4 x y)

\frac{\partial}{\partial x} (3 (x^{\frac{1}{3}} + y^{\frac{1}{3}}))

What is the integral from 0 to ln(5) of e^x(3-4e^x) ?
The integral from 0 to ln(5) of e^x(3-4e^x) is -36