How do you integrate $e^{- 1.5 x}$ $e^(-1.5x)$ ?

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How do you integrate $e^{- 1.5 x}$ $e^(-1.5x)$ ?

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Answer 1 · 2015-06-27T19:47:56

Use substitution with $u = - 1.5 x$ $u = -1.5x$

Explanation:

Any integral of the form $\int e^{k x} d x$ $int e^(kx) dx$ where $k$ $k$ is a constant, may be evaluated by using $u = k x$ $u = kx$ , so $d u = k d x$ $du = k dx$

$\int e^{k x} d x = \frac{1}{k} \int e^{k x} k d x$ $int e^(kx) dx = 1/k int e^(kx) kdx$

$= \frac{1}{k} \int e^{u} d u = \frac{1}{k} e^{u} + C$ $= 1/k int e^u du = 1/ke^u +C$

$= \frac{1}{k} e^{k x} + C$ $= 1/ke^(kx) +C$

This answer makes sense, because the derivative of $e^{k x}$ $e^(kx)$ is $e^{k x} \cdot k$ $e^(kx) *k$
That is: we know the integral of $e^{- 1.5 x}$ $e^(-1.5x)$ must involve $e^{- 1.5 x}$ $e^(-1.5x)$ , but the derivative of that is $\frac{d}{d x} (e^{- 1.5 x}) = - 1.5 e^{- 1.5 x}$ $d/dx(e^(-1.5x))= -1.5 e^(-1.5x)$ . We'll remove the $- 1.5$ $-1.5$ by dividing:

$\int e^{- 1.5 x} d x = \frac{1}{- 1.5} e^{- 1.5 x} + C$ $int e^(-1.5x) dx = 1/(-1.5) e^(-1.5x) +C$

Note: fraction with a decimal in the numerator or denominator looks strange to me. (I think: "Make up you mind! One or the other!")

$1.5 = \frac{3}{2}$ $1.5 = 3/2$ so $\frac{1}{1.5} = \frac{2}{3}$ $1/1.5 = 2/3$ and we can write our answer:

$- \frac{2}{3} e^{- 1.5 x} + C$ $-2/3 e^(-1.5x) +C$

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How do you integrate $e^{- 1.5 x}$ $e^(-1.5x)$ ?

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