Problem 36 - Calculus for Biology and Medicine

Question

(a) Use integration by parts to show that $$ \int x^{n} e^{x} d x=x^{n} e^{x}-n \int x^{n-1} e^{x} d x $$ Such formulas are called reduction formulas, since they reduce the exponent of \(x\) by 1 each time they are applied. (b) Apply the reduction formula in (a) repeatedly to compute \(\int x^{3} e^{x} d x\)

Step-by-Step Solution

Answer

\( \int x^3 e^x \, dx = x^3 e^x - 3x^2 e^x + 6x e^x - 6e^x + C \)

Step 1: Integration by Parts Formula

Recall the integration by parts formula: \( \int u \ dv = uv - \int v \ du \). This involves choosing appropriate functions for \( u \) and \( dv \).

Step 2: Choose Functions for Integration by Parts

For the integral \( \int x^n e^x \, dx \), let \( u = x^n \) and \( dv = e^x \, dx \). Then, differentiate and integrate to find \( du \) and \( v \): - \( du = n x^{n-1} \, dx \) - \( v = e^x \) since the integral of \( e^x \) is itself.

Step 3: Apply Integration by Parts Formula

Substitute \( u \), \( du \), and \( v \) into the integration by parts formula: \[ \int x^n e^x \, dx = x^n e^x - \int e^x (n x^{n-1}) \, dx \] This simplifies to: \[ \int x^n e^x \, dx = x^n e^x - n \int x^{n-1} e^x \, dx \] which is the reduction formula we wanted to show.

Step 4: Starting Reduction on \( \int x^3 e^x \, dx \)

Apply the reduction formula: \[ \int x^3 e^x \, dx = x^3 e^x - 3 \int x^2 e^x \, dx \] This reduces the power of \( x \) from 3 to 2 in the subsequent integral.

Step 5: Reduce the Integral \( \int x^2 e^x \, dx \)

Apply the reduction formula again: \[ \int x^2 e^x \, dx = x^2 e^x - 2 \int x^1 e^x \, dx \] This reduces the power of \( x \) from 2 to 1.

Step 6: Reduce the Integral \( \int x^1 e^x \, dx \)

Apply the reduction formula once more: \[ \int x^1 e^x \, dx = x^1 e^x - 1 \int x^0 e^x \, dx \] Which simplifies to: \[ \int x e^x \, dx = x e^x - \int e^x \, dx \]

Step 7: Solve the Final Integral \( \int e^x \, dx \)

Since the integral of \( e^x \) is \( e^x \), we have: \[ \int e^x \, dx = e^x \]

Step 8: Assemble Solutions for \( \int x^3 e^x \, dx \)

Combine the terms back together:- From Step 4: \( \int x^3 e^x \, dx = x^3 e^x - 3(x^2 e^x - 2(x e^x - e^x)) \)- Simplify and combine: - Final solution: \[ \int x^3 e^x \, dx = x^3 e^x - 3x^2 e^x + 6x e^x - 6e^x + C \]

Key Concepts

Integration by Parts

Integration by parts is a powerful technique in calculus that transforms the integral of a product of functions into an easier form to evaluate. Basically, it relies on the product rule of differentiation and is expressed as:

• \( \int u \ dv = uv - \int v \ du \)

The idea is to choose parts of the function to simplify the overall integration process. For example, if you have an integral \( \int x^n e^x \, dx \), you can assign different roles to parts of this expression:

• Let \( u = x^n \) which means the derivative \( du = nx^{n-1} \, dx \).
• Let \( dv = e^x \, dx \) so that \( v = e^x \), since the derivative of an exponential function \( e^x \) is \( e^x \).

By substituting these values into the integration by parts formula, the integral simplifies into something more manageable, resulting in the reduction formula: \[ \int x^n e^x \, dx = x^n e^x - n \int x^{n-1} e^x \, dx \] This is a significant step in solving complex integrals systematically.

Reduction Formulas

Reduction formulas are a clever way of solving integrals incrementally, and they are particularly beneficial when dealing with polynomial multiplied by exponential functions. The term "reduction" indicates that each application of the method reduces the problem's complexity.

In our exercise, using the reduction formula \[ \int x^n e^x \, dx = x^n e^x - n \int x^{n-1} e^x \, dx \]allows us to decrease the degree of the polynomial in each step. This stepwise reduction simplifies an otherwise tedious process of integration.

To illustrate, starting with \( \int x^3 e^x \, dx \), we apply the formula:

• Step 1: Results in \( x^3 e^x - 3 \int x^2 e^x \, dx \)
• Step 2: Further simplifies to \( x^3 e^x - 3(x^2 e^x - 2 \int x^1 e^x \, dx) \)
• Step 3: Continuing this way introduces no powers of \( x \), turning into simpler integrals like \( \int e^x \, dx \)

These steps consecutively lower the power of \( x \), effectively "reducing" the complexity of the evaluation. This systematic approach not only provides a neat solution but also makes heavy algebraic calculations manageable.

Exponential Functions

Exponential functions play a crucial role in calculus, particularly because they possess unique properties that make them easier to manipulate than other functions in integration. Most notably, the exponential function \( e^x \) remains unchanged when differentiated or integrated:

• The derivative of \( e^x \) with respect to \( x \) is \( e^x \).
• The integral of \( e^x \) is again \( e^x \).

These properties mean that exponential functions frequently appear in integration problems as they allow straightforward computations while interacting smoothly with other types of functions such as polynomials.

When using integration techniques like integration by parts and reduction formulas, exponential functions, such as \( e^x \), become invaluable. They simplify the process because their integrals do not become increasingly complicated as in the case of other functions. For integrals involving both polynomials and exponential functions, the combination can be systematically untangled using methods like reduction formulas. Each application of the formula breaks down the integral into simpler parts until only the basic integral of the exponential component remains. Thus, understanding exponential functions is essential to grasping more advanced integration techniques.