What is the integral from-1 to 5 of (3e^{-x})^2 ?

The integral from-1 to 5 of (3e^{-x})^2 is -9/(2e^{10)}+(9e^2)/2

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integral from-1 to 5 of (3e^{-x})^2

\int_{- 1}^{5} (3 e^{- x})^{2} d x

- \frac{9}{2 e ^{10}} + \frac{9 e ^{2}}{2}

Decimal Notation

33.25054 \dots

Solution steps

\int_{- 1}^{5} (3 e^{- x})^{2} d x

Apply Integration By Parts

: [9 e^{- 2 x} x - \int - 18 e^{- 2 x} x d x]_{- 1}^{5}

= [9 e^{- 2 x} x - \int - 18 e^{- 2 x} x d x]_{- 1}^{5}

\int - 18 e^{- 2 x} x d x = 9 e^{- 2 x} x + \frac{9}{2} e^{- 2 x}

= [9 e^{- 2 x} x - (9 e^{- 2 x} x + \frac{9}{2} e^{- 2 x})]_{- 1}^{5}

Simplify

= [- \frac{9}{2} e^{- 2 x}]_{- 1}^{5}

Compute the boundaries:

- \frac{9}{2 e ^{10}} + e^{2} \frac{9}{2}

= - \frac{9}{2 e ^{10}} + e^{2} \frac{9}{2}

Simplify

= - \frac{9}{2 e ^{10}} + \frac{9 e ^{2}}{2}

x \to 1 - lim (x^{(\frac{1}{1 - x})})

\frac{d}{d z} (x y e^{z})

f (x) = ln (x + 9)

\frac{d y}{d x} = e^{3 x} + 2 y

What is the integral from-1 to 5 of (3e^{-x})^2 ?
The integral from-1 to 5 of (3e^{-x})^2 is -9/(2e^{10)}+(9e^2)/2