Problem 30 - Calculus and its applications

Question

Evaluate using integration by parts or substitution. (Assume \(u>0\) in \(\ln\) u. Check by differentiating. $$ \int\left(x^{3}-x+1\right) e^{-x} d x $$

Step-by-Step Solution

Answer

Use integration by parts to find \( \int (x^3-x+1)e^{-x} \, dx = -(x^3-x+1)e^{-x} - 3(x^2+2x+2)e^{-x} + C \). Verify by differentiation.

Step 1: Identify the Technique

We are given the integral \( \int (x^3 - x + 1)e^{-x} \, dx \). This integral can be solved using integration by parts since it involves a polynomial \((x^3 - x + 1)\) and an exponential function \(e^{-x}\) that can be integrated easily.

Step 2: Choose \(u\) and \(dv\)

In integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). Choose \( u = x^3 - x + 1 \) and \( dv = e^{-x} \, dx \).

Step 3: Differentiate and Integrate

Differentiate \( u \) to find \( du \):\[ du = (3x^2 - 1) \, dx \]. Integrate \( dv \) to find \( v \):\[ v = \int e^{-x} \, dx = -e^{-x} \].

Step 4: Apply the Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \[ \int (x^3 - x + 1) e^{-x} \, dx = -(x^3 - x + 1)e^{-x} - \int (-e^{-x}) (3x^2 - 1) \, dx \].

Step 5: Simplify the New Integral

The integral becomes: \[ -(x^3 - x + 1)e^{-x} + \int (3x^2 - 1) e^{-x} \, dx \]. Here, the process of integration by parts must be repeated on \( \int 3x^2 e^{-x} \, dx \) and \( \int e^{-x} \, dx \).

Step 6: Repeat Integration by Parts

For \( \int 3x^2 e^{-x} \, dx \), use integration by parts again with \( u = 3x^2 \), \( dv = e^{-x} \, dx \). Then, continue integration by parts stepwise to eventually solve for the entire expression.

Step 7: Simplify and Combine Results

After applying integration by parts multiple times and simplifying, you will reach the solution to the integral. Thorough simplification should yield:\[ \int (x^3 - x + 1) e^{-x} \, dx = -(x^3 - x + 1)e^{-x} - 3(x^2 + 2x + 2)e^{-x} + C \],where \( C \) is the constant of integration.

Step 8: Verify by Differentiation

Differentiate the result to verify that it brings us back to the original integrand \((x^3-x+1)e^{-x}\). Confirm that every term aligns to ensure the calculation is correct.

Key Concepts

Integral Calculus

Integral Calculus is a subfield of calculus that focuses on the concept of integration and the accumulation of quantities. Integration allows us to find areas under curves, solve differential equations, and calculate total quantities when only rates of change are known.
When working with integrals, we essentially perform the reverse operation of differentiation. The goal is to find the function whose derivative gives us the original function we started with.

In this exercise, the technique used is **Integration by Parts**. This method is derived from the product rule for differentiation and is particularly useful when dealing with integrals of products of functions, like polynomial and exponential functions in this case.

• The formula for integration by parts is derived from: \( \int u \, dv = uv - \int v \, du \)
• Choosing which part of the expression to differentiate (\(u\)) and which to integrate (\(dv\)) is key to simplifying the equation.

In our problem, by carefully selecting \(u = x^3 - x + 1\) (the polynomial) and \(dv = e^{-x} \, dx\) (the exponential function), we simplify the integration process. Repeated application of integration by parts resolves the integral.

Exponential Functions

Exponential functions are a vital part of mathematics, especially in calculus, due to their unique properties. They are expressed in the form \(e^{x}\), where \(e\) is a constant approximately equal to 2.71828.
These functions grow rapidly and are widely used in various fields, including finance, physics, and biology. Their significance lies in their growth patterns that can model real-world phenomena effectively.

Integrating exponential functions is relatively straightforward, as they often retain their form. For example, the integral of \(e^{-x}\) is simply \(-e^{-x}\). This makes them quite manageable within integration by parts.

• Exponential functions often appear as a part of the product in complex integrals, requiring techniques like integration by parts to simplify and solve.
• In the exercise, the exponential function \(e^{-x}\) was chosen as \(dv\) to easily find its integral.

By understanding how exponential functions behave, we can simplify the integration processes that involve them, ideal for scenarios like the one in this exercise.

Polynomial Functions

Polynomial functions are expressions that involve sums of powers of a variable, typically expressed as \(a_nx^n + a_{n-1}x^{n-1} + \, ... \, + a_1x + a_0\).
They are fundamental in mathematics as they provide a framework for expressing complex relationships in a simplified manner. Integrals of polynomial functions are generally more straightforward because they involve simple algebraic manipulation.

• With polynomials, integration corresponds to adding one to the power of each term and dividing by the new power, except when mixed with other function types.
• The factorial increase in complexity, when combined with non-polynomial functions like exponentials, necessitates advanced methods such as integration by parts.

In the exercise, the polynomial \(x^3 - x + 1\) was straightforward to differentiate, but challenging to integrate directly as part of the product with \(e^{-x}\). That is why integration by parts is employed.
Recognizing and handling polynomial functions within integrals is essential for solving complex differential equations and modeling dynamic systems.