Problem 37 - Calculus for Biology and Medicine

Question

(a) Use integration by parts to verify the validity of the reduction formula $$ \int x^{n} e^{a x} d x=\frac{1}{a} x^{n} e^{e x}-\frac{n}{a} \int x^{n-1} e^{a x} d x $$ where \(a\) is a constant not equal to 0 . (b) Apply the reduction formula in (a) to compute \(\int x^{2} e^{-3 x} d x\).

Step-by-Step Solution

Answer

The reduction formula is verified, and the integral is \(-\frac{1}{3} x^2 e^{-3x} - \frac{2}{9} x e^{-3x} + \frac{2}{27} e^{-3x} + C\)."

Step 1: Setup for Integration by Parts

To use integration by parts, we need to identify parts of the integrand to differentiate and integrate. Let \( u = x^n \) and \( dv = e^{ax} \, dx \). Then, \( du = n x^{n-1} \, dx \) and \( v = \frac{1}{a} e^{ax} \). The integration by parts formula \( \int u \, dv = uv - \int v \, du \) will be used.

Step 2: Apply Integration by Parts Formula

Substitute the parts into the integration by parts formula: \( \int x^n e^{ax} \, dx = x^n \cdot \frac{1}{a} e^{ax} - \int \frac{1}{a} e^{ax} \cdot n x^{n-1} \, dx \). Simplify this expression to get \( \frac{1}{a} x^n e^{ax} - \frac{n}{a} \int x^{n-1} e^{ax} \, dx \), which matches the given reduction formula.

Step 3: Verify the Reduction Formula

Since applying integration by parts led to the expression \( \int x^n e^{ax} \, dx = \frac{1}{a} x^n e^{ax} - \frac{n}{a} \int x^{n-1} e^{ax} \, dx \), the reduction formula is verified to be correct.

Step 4: Apply Reduction Formula for \(n = 2\) and \(a = -3\)

Plug \( n = 2 \) and \( a = -3 \) into the reduction formula: \( \int x^2 e^{-3x} \, dx = \frac{1}{-3} x^2 e^{-3x} - \frac{2}{-3} \int x e^{-3x} \, dx \). Simplify to get \( -\frac{1}{3} x^2 e^{-3x} + \frac{2}{3} \int x e^{-3x} \, dx \).

Step 5: Use Reduction Formula Again with \(n = 1\)

Now apply the formula again for \( \int x e^{-3x} \, dx \) with \( n = 1 \): \( \int x e^{-3x} \, dx = \frac{1}{-3} x e^{-3x} - \frac{1}{-3} \int e^{-3x} \, dx \). This simplifies to \( -\frac{1}{3} x e^{-3x} + \frac{1}{3} \int e^{-3x} \, dx \).

Step 6: Integrate the Remaining Term

Calculate \( \int e^{-3x} \, dx = -\frac{1}{3} e^{-3x} + C \), since the integral of \( e^{ax} \) is \( \frac{1}{a} e^{ax} + C \).

Step 7: Substitute Back and Final Calculation

Substitute back \( \int x e^{-3x} \, dx \) into the main expression: \( \int x e^{-3x} \, dx = -\frac{1}{3} x e^{-3x} + \frac{1}{9} e^{-3x} \). Then, substitute this into the \( \int x^2 e^{-3x} \, dx \) to get \( -\frac{1}{3} x^2 e^{-3x} + \frac{2}{3} \left(-\frac{1}{3} x e^{-3x} + \frac{1}{9} e^{-3x} \right) \).

Step 8: Simplify Final Expression

Simplify to get \( -\frac{1}{3} x^2 e^{-3x} - \frac{2}{9} x e^{-3x} + \frac{2}{27} e^{-3x} + C \). This is the evaluated integral \( \int x^2 e^{-3x} \, dx \).

Key Concepts

Reduction Formula

Reduction formulas are a powerful tool in calculus that allow us to simplify the process of solving complex integrals. They work by introducing a recurrence relation to break down a function into simpler parts. In general, a reduction formula takes an integral of higher complexity and reduces it to a simpler integral that resembles the original one. This makes it easier to handle.

The reduction formula used in this exercise is:

• \[\int x^n e^{a x} \, dx = \frac{1}{a} x^n e^{a x} - \frac{n}{a} \int x^{n-1} e^{a x} \, dx\]

Here, it's clear that instead of directly solving the integral \(\int x^n e^{a x} \, dx\), we reduce it to a simpler form involving an integral of \(x^{n-1} e^{a x}\). This significantly simplifies the integration process, particularly with exponential functions, where directly integrating can be challenging.

To construct these formulas, integration by parts is often used. Each application of the formula reduces the complexity by decreasing the power of \(x\), making it increasingly manageable. This recursive approach turns many challenging integrals into a series of easier steps.

Calculus Problem Solving

Solving calculus problems often involves breaking them down into smaller, more manageable parts. The integration by parts technique is particularly handy for this, as it leverages the product rule for differentiation in reverse. The integration by parts formula is:

• \[\int u \, dv = uv - \int v \, du\]

In this problem, identifying the parts \(u\) and \(dv\) correctly is crucial. For this exercise:

• Choose \(u = x^n\), which simplifies to \(du = n x^{n-1} \, dx\).
• Choose \(dv = e^{ax} \, dx\), leading to \(v = \frac{1}{a} e^{ax}\).

Using these parts in the integration by parts formula makes solving complex integrals more systematic. By repeatedly applying these principles, as seen in the step-by-step solution, more complicated expressions can be solved with ease. Dividing the integration process into sequential steps helps in clarifying the task and transforming it into a series of simple operations.

By practicing these methods, students build a robust understanding of the mechanics of calculus and get better equipped to tackle more complicated calculus problems.

Exponential Functions

Exponential functions have the form \(e^{ax}\) and exhibit unique properties that differentiate them from other functions. One such property is that the derivative and the integral of \(e^{ax}\) are both closely related to the function itself. The derivative of \(e^{ax}\) is \(ae^{ax}\), and its integral \(\int e^{ax} \, dx = \frac{1}{a}e^{ax} + C\).

These properties make exponential functions particularly useful in the reduction formula discussed. Working with exponential functions frequently involves factors of \(e^{ax}\) that appear both in numerator and denominator, facilitating simplification.
The exercise demonstrates how exponential functions paired with polynomial terms \(x^n\) can increase the complexity of integration. However, with the clever use of exponential properties and strategies like reduction formulas, such integrals become easier to manage.Understanding these characteristics is essential for mastering topics in calculus. Exponential functions play a significant role in countless mathematical models, making them vital for students aiming to excel in applied mathematics and beyond. By deeply understanding their behavior and application in integrals, students can better handle various advanced mathematical problems.