Question

Find the integral by using the simplest method. Not all problems require integration by parts. $$ \int x^{2} e^{-x} d x $$

Short answer

The integral is \( -e^{-x} (x^2 + 2x + 2) + C \).

Step-by-step solution

Identify the Method

To solve the integral \( \int x^2 e^{-x} \, dx \), we first identify the method to use. In this case, integration by parts is appropriate, as it involves a product of a polynomial \( x^2 \) and an exponential function \( e^{-x} \).

Apply Integration by Parts Formula

Recall the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). We choose \( u = x^2 \) and \( dv = e^{-x} \, dx \). Then, we need to find \( du \) and \( v \).

Differentiate and Integrate

Differentiate \( u = x^2 \) to get \( du = 2x \, dx \). Integrate \( dv = e^{-x} \, dx \) to get \( v = -e^{-x} \).

Substitute and Simplify

Substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula: \( \int x^2 e^{-x} \, dx = -x^2 e^{-x} - \int -2x e^{-x} \, dx \). Simplify to \( \int x^2 e^{-x} \, dx = -x^2 e^{-x} + 2\int x e^{-x} \, dx \).

Repeat Integration by Parts

For the integral \( \int x e^{-x} \, dx \), apply integration by parts again. Choose \( u = x \) and \( dv = e^{-x} \, dx \). Then \( du = dx \) and \( v = -e^{-x} \).

Substitute and Simplify Again

Substitute in the formula: \( \int x e^{-x} \, dx = -x e^{-x} - \int -e^{-x} \, dx \). Simplify to \( \int x e^{-x} \, dx = -x e^{-x} + \int e^{-x} \, dx \), which becomes \( -x e^{-x} - e^{-x} \).

Final Integration and Substitution

Substitute \( \int x e^{-x} \, dx = -x e^{-x} - e^{-x} \) back into the expression from step 4: \( \int x^2 e^{-x} \, dx = -x^2 e^{-x} + 2(-x e^{-x} - e^{-x}) \).

Simplify the Overall Expression

Multiply and simplify: \( \int x^2 e^{-x} \, dx = -x^2 e^{-x} - 2x e^{-x} - 2e^{-x} \). Factor out \(-e^{-x}\): \( \int x^2 e^{-x} \, dx = -e^{-x} (x^2 + 2x + 2) \).

Add the Constant of Integration

Since we are dealing with an indefinite integral, remember to include the constant of integration \( C \) at the end: \( \int x^2 e^{-x} \, dx = -e^{-x} (x^2 + 2x + 2) + C \).

Key concepts

Definite Integrals

Definite integrals differ from indefinite integrals in that they calculate the area under a curve within specific limits. With definite integrals, you determine a numerical value that represents this area.
When you perform a definite integral, it helps to think of it as summing up an infinite number of small rectangles under the curve. This is fundamentally the idea of integration in calculus.
In the context of integration by parts, understanding definite integrals can be crucial, as this method can help find exact values by evaluating the integral at upper and lower limits. Regular practice is essential to understand each step thoroughly and avoid missing the bounds of integration.

Exponential Functions

Exponential functions, like the function in the exercise, have the general form of \( e^{x} \), where \( e \) is the base of the natural logarithm. These functions grow or decay rapidly, which is why they often appear in speed-related problems or naturally occurring processes.
For instance, the integral \( \int e^{-x} dx \) involves an exponential decay because of the negative exponent. This characteristic makes exponential functions versatile and frequently tackled in calculus.

• In integration problems, exponential functions are regular components, often paired with polynomials or trigonometric functions.
• When integrating an exponential function, it often results in itself, scaled by a constant.

Understanding exponential behavior helps with anticipating the outcome and checking the results. Practice identifying exponential functions and applying appropriate integration techniques.

Polynomials

Polynomials are algebraic expressions that consist of variables raised to whole-number powers. The expression \( x^2 \) is a simple polynomial with a degree of 2.

• Polyonomials are versatile in mathematics, appearing in statistics, physics, and economics to model a variety of real-world phenomena.
• Understanding how to differentiate and integrate polynomials is a vital skill in calculus, as these operations form the basics of many other complex problems.
• When integrating combinations of polynomials like \( x^2 \, e^{-x} \), it's essential to manage the polynomial by reducing its degree with each successful integration step.

In integration problems, polynomials like \( x^2 \) require careful choice of methods like integration by parts, especially when combined with functions that need alternative approaches. Mastery of polynomial manipulation can significantly streamline the integration process.