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(a) \( \int x^{2} e^{-2 x} d x \)

Ask by Lindsey Olson. in Iraq
Feb 03,2025

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Tutor-Verified Answer

Answer

\[ \int x^{2} e^{-2x} dx = -\frac{x^{2}}{2e^{2x}} - \frac{x}{2e^{2x}} - \frac{1}{4e^{2x}} + C \]

Solution

Calculate or simplify the expression \( \int x^{2} e^{-2x} dx \). Evaluate the integral by following steps: - step0: Evaluate using partial integration formula: \(\int x^{2}e^{-2x} dx\) - step1: Prepare for integration by parts: \(\begin{align}&u=x^{2}\\&dv=e^{-2x}dx\end{align}\) - step2: Calculate the derivative: \(\begin{align}&du=2xdx\\&dv=e^{-2x}dx\end{align}\) - step3: Evaluate the integral: \(\begin{align}&du=2xdx\\&v=-\frac{1}{2e^{2x}}\end{align}\) - step4: Substitute the values into formula: \(x^{2}\left(-\frac{1}{2e^{2x}}\right)-\int 2x\left(-\frac{1}{2e^{2x}}\right) dx\) - step5: Calculate: \(-\frac{x^{2}}{2e^{2x}}-\int -\frac{x}{e^{2x}} dx\) - step6: Use properties of integrals: \(-\frac{x^{2}}{2e^{2x}}+\int \frac{x}{e^{2x}} dx\) - step7: Evaluate the integral: \(-\frac{x^{2}}{2e^{2x}}-\frac{x}{2e^{2x}}-\frac{1}{4e^{2x}}\) - step8: Add the constant of integral C: \(-\frac{x^{2}}{2e^{2x}}-\frac{x}{2e^{2x}}-\frac{1}{4e^{2x}} + C, C \in \mathbb{R}\) النتيجة هي: \[ \int x^{2} e^{-2x} dx = -\frac{x^{2}}{2e^{2x}} - \frac{x}{2e^{2x}} - \frac{1}{4e^{2x}} + C \] حيث \(C\) هو ثابتintegration.

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Beyond the Answer

To tackle the integral \( \int x^{2} e^{-2 x} d x \), we can use the method of integration by parts. Let's set \( u = x^2 \) and \( dv = e^{-2x} dx \). This gives us \( du = 2x dx \) and \( v = -\frac{1}{2} e^{-2x} \). After applying integration by parts, we get: \[ \int x^{2} e^{-2 x} d x = -\frac{1}{2} x^2 e^{-2x} - \int -\frac{1}{2} (2x) e^{-2x} dx. \] This leads us to a new integral that we can simplify and solve using the same strategy until we reach a manageable result. Keep going with integration by parts until you have an integral that you can solve easily or recognize a pattern from.

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