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

The integral from 1 to 3 of ((x)^3+e^{-3x}) is 20+(e^6-1)/(3e^9)

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

\int_{1}^{3} ((x)^{3} + e^{- 3 x}) d x

20 + \frac{e ^{6} - 1}{3 e ^{9}}

Solution steps

\int_{1}^{3} x^{3} + e^{- 3 x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{3} x^{3} d x + \int_{1}^{3} e^{- 3 x} d x

\int_{1}^{3} x^{3} d x = 20

\int_{1}^{3} e^{- 3 x} d x = \frac{e ^{6} - 1}{3 e ^{9}}

= 20 + \frac{e ^{6} - 1}{3 e ^{9}}

x \to 36 lim (x)

x \to 3 lim ((4 x + 5) (3 x - 2))

x \to - \infty lim (1 + x^{2})

y = x ln (x)

f (x) = 3 x, at x = 16

What is the integral from 1 to 3 of ((x)^3+e^{-3x}) ?
The integral from 1 to 3 of ((x)^3+e^{-3x}) is 20+(e^6-1)/(3e^9)