x.e−x integral | What is integral of x.e^-x? | ∫x.e−x.dx=?

Understanding the Integral of ( x \cdot e^{-x} )

Calculating the integral ( \int x \cdot e^{-x} , dx ) may seem daunting at first glance, but with the application of integration techniques such as integration by parts, we can arrive at a solution. This integral combines polynomial and exponential functions, giving it unique properties that are useful in various fields, including physics and engineering.

Step-by-Step Solution

To evaluate ( \int x \cdot e^{-x} , dx ), we will use integration by parts, a method that leverages the product rule of differentiation. The formula for integration by parts is:

$\int u , dv = uv - \int v , du$

In this case, we can let:

( u = x ) → which implies ( du = dx )

( dv = e^{-x} , dx ) → which implies ( v = -e^{-x} )

Applying Integration by Parts

Substituting these choices into the integration by parts formula:

$\int x e^{-x} , dx = uv - \int v , du$

$= x(-e^{-x}) - \int (-e^{-x}) , dx$

$= -x e^{-x} + \int e^{-x} , dx$

Next, we need to compute ( \int e^{-x} , dx ):

$\int e^{-x} , dx = -e^{-x}$

Substituting this back into our previous result:

$\int x e^{-x} , dx = -x e^{-x} - e^{-x} + C$

Where ( C ) is the constant of integration.

Final Result

Combining the terms, we arrive at the final solution:

$\int x e^{-x} , dx = -e^{-x}(x + 1) + C$

Applications and Importance

The integral ( \int x e^{-x} , dx ) is particularly significant in probability theory, specifically in calculating moments of certain distributions like the exponential distribution. The presence of ( e^{-x} ) in the integral signifies damping or decay, which is observed in processes described by exponential functions.

Moreover, this integral serves as a standard example in calculus courses, demonstrating the application of integration by parts. Understanding how to tackle such integrals is crucial for students and professionals in mathematics, physics, engineering, and economics.

Conclusion

In summary, the integral ( \int x e^{-x} , dx ) can be elegantly solved using integration by parts, yielding the result ( -e^{-x}(x + 1) + C ). This example showcases not only a fundamental calculus technique but also the broader applicability of integrals involving exponential functions in various scientific fields.