Question
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Answer
The integral of \( x e^{-2x} \) with respect to \( x \) is \( -\frac{x}{2e^{2x}} - \frac{1}{4e^{2x}} + C \).
Solution
Calculate the integral \( \int x e^{-2x} dx \).
Evaluate the integral by following steps:
- step0: Evaluate using partial integration formula:
\(\int xe^{-2x} dx\)
- step1: Prepare for integration by parts:
\(\begin{align}&u=x\\&dv=e^{-2x}dx\end{align}\)
- step2: Calculate the derivative:
\(\begin{align}&du=dx\\&dv=e^{-2x}dx\end{align}\)
- step3: Evaluate the integral:
\(\begin{align}&du=dx\\&v=-\frac{1}{2e^{2x}}\end{align}\)
- step4: Substitute the values into formula:
\(x\left(-\frac{1}{2e^{2x}}\right)-\int 1\times \left(-\frac{1}{2e^{2x}}\right) dx\)
- step5: Calculate:
\(-\frac{x}{2e^{2x}}-\int -\frac{1}{2e^{2x}} dx\)
- step6: Use properties of integrals:
\(-\frac{x}{2e^{2x}}+\int \frac{1}{2e^{2x}} dx\)
- step7: Evaluate the integral:
\(-\frac{x}{2e^{2x}}-\frac{1}{4e^{2x}}\)
- step8: Add the constant of integral C:
\(-\frac{x}{2e^{2x}}-\frac{1}{4e^{2x}} + C, C \in \mathbb{R}\)
The integral of \(x e^{-2x}\) with respect to \(x\) is given by:
\[ \int x e^{-2x} dx = -\frac{x}{2e^{2x}} - \frac{1}{4e^{2x}} + C \]
where \(C\) is the constant of integration.
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Simplify this solution Mind Expander
To solve the integral \( \int x e^{-2 x} d x \), we can use integration by parts. Let's set \( u = x \) (then \( du = dx \)) and \( dv = e^{-2x} dx \) (which gives us \( v = -\frac{1}{2} e^{-2x} \)). Applying integration by parts, we have: \[ \int x e^{-2 x} d x = uv - \int v du \] Substituting our values in, we get: \[ = -\frac{1}{2} x e^{-2x} - \int -\frac{1}{2} e^{-2x} dx \] Now, solving the remaining integral: \[ = -\frac{1}{2} x e^{-2x} + \frac{1}{4} e^{-2x} + C \] Thus, the final result is: \[ \int x e^{-2 x} d x = -\frac{1}{2} x e^{-2x} + \frac{1}{4} e^{-2x} + C \]
