Integral of x2/e?: Techniques and Applications
Understanding the integration of functions, especially those involving exponential functions, is a fundamental skill in calculus. This article delves into the integral of x2/ex, exploring techniques such as integration by parts to solve the problem.
Introduction to Integration by Parts
Integration by parts is a powerful technique for integrating products of functions. The formula for integration by parts is given by:
[ int u , dv uv - int v , du ]This formula is derived from the product rule of differentiation. By applying this technique step-by-step, we can integrate complex functions such as x2/ex.
Step-by-Step Solution
Step 1: First Application of Integration by Parts
To start, let's set:
[ u x^2 quad Rightarrow quad du 2x , dx ] [ dv frac{1}{e^x} , dx quad Rightarrow quad v -frac{1}{e^x} ]Applying the integration by parts formula:
[ int frac{x^2}{e^x} , dx -frac{x^2}{e^x} - int -frac{1}{e^x} cdot 2x , dx ]This simplifies to:
[ -frac{x^2}{e^x} 2 int frac{x}{e^x} , dx ]Step 2: Second Application of Integration by Parts
Next, we need to evaluate (int frac{x}{e^x} , dx). Again, using integration by parts:
[ u x quad Rightarrow quad du dx ] [ dv frac{1}{e^x} , dx quad Rightarrow quad v -frac{1}{e^x} ]Applying the integration by parts formula:
[ int frac{x}{e^x} , dx -frac{x}{e^x} - int -frac{1}{e^x} , dx ]This simplifies to:
[ -frac{x}{e^x} int frac{1}{e^x} , dx ]The integral of (frac{1}{e^x}) is (-frac{1}{e^x}), so:
[ int frac{x}{e^x} , dx -frac{x}{e^x} frac{1}{e^x} ]Step 3: Combining Results
Substitute the result from Step 2 back into the equation from Step 1:
[ int frac{x^2}{e^x} , dx -frac{x^2}{e^x} 2 left( -frac{x}{e^x} frac{1}{e^x} right) ]This simplifies to:
[ int frac{x^2}{e^x} , dx -frac{x^2}{e^x} - frac{2x}{e^x} frac{2}{e^x} ]Combining like terms, we get:
[ int frac{x^2}{e^x} , dx -frac{x^2 2x 2}{e^x} C ]Where C is the constant of integration.
Generalization Using a Polynomial
For a more general approach, consider the integral of a polynomial times an exponential function. Let:
[ int p(x)e^{-x} , dx ]where p(x) is a polynomial. By using the method of integration by parts and induction, we can find that:
[ int p(x)e^{-x} , dx q(x)e^{-x} C ]Where q(x) is a polynomial of the same degree as p(x). This general fact can be proven by induction on the degree of the polynomial.
Conclusion
Understanding the integral of x2/ex not only reinforces the method of integration by parts but also provides a valuable technique for solving more complex integrals involving exponential functions. Whether you are a student, mathematician, or engineer, mastering these techniques can significantly enhance your problem-solving skills.