What is the integral from-1 to 1 of ((x+3)^3e^x)/2 ?

The integral from-1 to 1 of ((x+3)^3e^x)/2 is (34e^2-2)/(2e)

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integral from-1 to 1 of ((x+3)^3e^x)/2

\int_{- 1}^{1} \frac{( x + 3 ) ^{3} e ^{x}}{2} d x

\frac{34 e ^{2} - 2}{2 e}

Decimal

45.84291 \dots

Solution steps

\int_{- 1}^{1} \frac{( x + 3 ) ^{3} e ^{x}}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- 1}^{1} (x + 3)^{3} e^{x} d x

Expand

(x + 3)^{3} e^{x} : e^{x} x^{3} + 9 e^{x} x^{2} + 27 e^{x} x + 27 e^{x}

= \frac{1}{2} \cdot \int_{- 1}^{1} e^{x} x^{3} + 9 e^{x} x^{2} + 27 e^{x} x + 27 e^{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

= \frac{1}{2} (\int_{- 1}^{1} e^{x} x^{3} d x + \int_{- 1}^{1} 9 e^{x} x^{2} d x + \int_{- 1}^{1} 27 e^{x} x d x + \int_{- 1}^{1} 27 e^{x} d x)

\int_{- 1}^{1} e^{x} x^{3} d x = - 2 e + \frac{16}{e}

\int_{- 1}^{1} 9 e^{x} x^{2} d x = 9 (e - \frac{5}{e})

\int_{- 1}^{1} 27 e^{x} x d x = \frac{54}{e}

\int_{- 1}^{1} 27 e^{x} d x = 27 (e - \frac{1}{e})

= \frac{1}{2} (- 2 e + \frac{16}{e} + 9 (e - \frac{5}{e}) + \frac{54}{e} + 27 (e - \frac{1}{e}))

Simplify

\frac{1}{2} (- 2 e + \frac{16}{e} + 9 (e - \frac{5}{e}) + \frac{54}{e} + 27 (e - \frac{1}{e})) : \frac{34 e ^{2} - 2}{2 e}

= \frac{34 e ^{2} - 2}{2 e}

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

\int_{0}^{2} x^{2} 2 x - x^{2} d x

\int_{0}^{5} 8 e^{2 x} d x

\int_{x}^{2 x} e^{- t^{2}} d t

\int_{0}^{2} \frac{3 x}{( x + 1 ) ^{3}} d x

What is the integral from-1 to 1 of ((x+3)^3e^x)/2 ?
The integral from-1 to 1 of ((x+3)^3e^x)/2 is (34e^2-2)/(2e)