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\( \left(1.0\right. \) punto) \( \int_{1}^{2} 2 x e^{-3 x} d x \)

Ask by Coleman Clark. in Colombia
Feb 02,2025

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Answer

La integral definida \( \int_{1}^{2} 2x\, e^{-3x}\, dx \) es aproximadamente igual a 0.040368.

Solution

Para calcular la integral definida \[ \int_{1}^{2} 2x\, e^{-3x}\, dx, \] utilizaremos el método de **integración por partes**. Recordemos que la fórmula de integración por partes es: \[ \int u\, dv = uv - \int v\, du. \] ### Paso 1: Identificar \( u \) y \( dv \) Elegimos: \[ u = 2x \quad \Rightarrow \quad du = 2\, dx \] \[ dv = e^{-3x}\, dx \quad \Rightarrow \quad v = \int e^{-3x}\, dx = -\frac{1}{3} e^{-3x} \] ### Paso 2: Aplicar la fórmula de integración por partes Sustituyendo en la fórmula: \[ \int 2x\, e^{-3x}\, dx = uv - \int v\, du \] \[ = \left(2x\right) \left(-\frac{1}{3} e^{-3x}\right) - \int \left(-\frac{1}{3} e^{-3x}\right) \left(2\, dx\right) \] \[ = -\frac{2}{3} x\, e^{-3x} + \frac{2}{3} \int e^{-3x}\, dx \] ### Paso 3: Resolver la integral restante \[ \int e^{-3x}\, dx = -\frac{1}{3} e^{-3x} + C \] Sustituyendo de nuevo: \[ - \frac{2}{3} x\, e^{-3x} + \frac{2}{3} \left( -\frac{1}{3} e^{-3x} \right) + C \] \[ = -\frac{2}{3} x\, e^{-3x} - \frac{2}{9} e^{-3x} + C \] ### Paso 4: Evaluar la integral definida Evaluamos entre los límites de integración \( x = 1 \) y \( x = 2 \): \[ \left[ -\frac{2}{3} x\, e^{-3x} - \frac{2}{9} e^{-3x} \right]_{1}^{2} = \left( -\frac{2}{3} \cdot 2\, e^{-6} - \frac{2}{9} e^{-6} \right) - \left( -\frac{2}{3} \cdot 1\, e^{-3} - \frac{2}{9} e^{-3} \right) \] \[ = \left( -\frac{4}{3} e^{-6} - \frac{2}{9} e^{-6} \right) - \left( -\frac{2}{3} e^{-3} - \frac{2}{9} e^{-3} \right) \] \[ = -\frac{14}{9} e^{-6} + \frac{8}{9} e^{-3} \] \[ = \frac{8}{9} e^{-3} - \frac{14}{9} e^{-6} \] ### Paso 5: Calcular el valor numérico aproximado Calculando los valores aproximados: \[ e^{-3} \approx 0.04978707 \] \[ e^{-6} \approx 0.002478752 \] Sustituyendo: \[ \frac{8}{9} \cdot 0.04978707 - \frac{14}{9} \cdot 0.002478752 \approx 0.0442326 - 0.0038647 \approx 0.040368 \] ### Resultado Final \[ \int_{1}^{2} 2x\, e^{-3x}\, dx \approx 0.040368 \]

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The Deep Dive

To solve the integral \( \int_{1}^{2} 2 x e^{-3 x} d x \), we can use integration by parts. Set \( u = 2x \) and \( dv = e^{-3x} dx \). Then, calculate \( du = 2 dx \) and \( v = -\frac{1}{3} e^{-3x} \). Using integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get: \[ \int 2x e^{-3x} dx = -\frac{2}{3} x e^{-3x} + \frac{2}{9} e^{-3x} + C \] Now, evaluate this from \( x = 1 \) to \( x = 2 \): \[ \left[-\frac{2}{3} \cdot 2 e^{-6} + \frac{2}{9} e^{-6}\right] - \left[-\frac{2}{3} \cdot 1 e^{-3} + \frac{2}{9} e^{-3}\right] \] Calculating those values gives you the definite integral result effectively!

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