Problem 12 - Thomas Calculus

Question

Evaluate the integrals. \(\int p^{4} e^{-p} d p\)

Step-by-Step Solution

Answer

The evaluated integral is \(-p^4 e^{-p} + 4p^3 e^{-p} - 12p^2 e^{-p} + 24p e^{-p} - 24e^{-p} + C\).

Step 1: Choose the Integration Technique

To solve the integral \( \int p^{4} e^{-p} dp \), we will use integration by parts. Integration by parts is useful when we have a product of two functions where one function can be easily differentiated, and the other can be easily integrated.

Step 2: Set Up Integration by Parts

Recall the formula for integration by parts: \( \int u \, dv = uv - \int v \, du \). For this problem, choose \( u = p^4 \) and \( dv = e^{-p} \, dp \). The function \( u \) will get simplified by differentiating, while \( dv \) is easily integrable.

Step 3: Differentiate and Integrate

Differentiate \( u = p^4 \) to get \( du = 4p^3 \, dp \). Integrate \( dv = e^{-p} \, dp \) to obtain \( v = -e^{-p} \).

Step 4: Apply Integration by Parts Formula

Substitute into the integration by parts formula: \( \int p^4 e^{-p} \, dp = -p^4 e^{-p} - \int (-e^{-p}) (4p^3) \, dp \). Simplify this to \( -p^4 e^{-p} + 4 \int p^3 e^{-p} \, dp \).

Step 5: Repeat Integration by Parts

For the term \( \int p^3 e^{-p} \, dp \), apply integration by parts again. Choose \( u = p^3 \) and \( dv = e^{-p} \, dp \). Then, \( du = 3p^2 \, dp \) and \( v = -e^{-p} \).

Step 6: Perform Further Calculations

Apply the integration by parts formula to \( \int p^3 e^{-p} dp \): Substituting the terms, we have \( -p^3 e^{-p} + 3 \int p^2 e^{-p} dp \). Repeat this process for each resulting integral, reducing the power of \( p \) each time.

Step 7: Solve Remaining Integrals

Continue using integration by parts in the same manner until you resolve \( \int p^2 e^{-p} dp \), \( \int p e^{-p} dp \), and \( \int e^{-p} dp \). Each iteration decreases the power of \( p \) until the recursion stops with the integral of \( e^{-p} \).

Step 8: Compile the Solution

Combine all parts from each step, simplifying as you go along. After resolving each integral, the original integral simplifies to a combination of terms like \(-p^4 e^{-p} + 4(p^3 e^{-p} - 3p^2 e^{-p} + 6p e^{-p} - 6e^{-p}) + C\), with \( C \) as the constant of integration.

Key Concepts

Integration Techniques

Understanding integration techniques is vital when tackling complex integrals like \(\int p^4 e^{-p} \, dp\). One powerful method is **integration by parts**. This technique is the counterpart to the product rule for differentiation and is especially useful for integrals of products of functions.
We specifically apply it when dealing with a product of two functions where:

• One function simplifies upon differentiation, and
• The other is straightforward to integrate.

The formula for integration by parts is \(\int u \, dv = uv - \int v \, du\). Choosing the right \(u\) and \(dv\) is crucial for simplifying the problem. In our example, setting \(u = p^4\) simplifies each time we differentiate. Meanwhile, \(dv = e^{-p} \, dp\) is an exponential function that remains manageable when integrated.

Definite and Indefinite Integrals

Integrals come in two forms: definite and indefinite. Indefinite integrals, like \(\int p^4 e^{-p} \, dp\), do not have specified limits and yield a general solution involving a constant of integration \(C\). This constant arises because differentiation of constants gives zero, but in an indefinite integral, you need to account for any arbitrary constant.
Unlike indefinite integrals, **definite integrals** have upper and lower bounds. They provide the net area between the graph of a function and the \(x\)-axis, within the specified limits. To compute a definite integral after obtaining the antiderivative, simply substitute the bounds into the result and subtract.

Exponential Functions

Exponential functions, like \(e^{-p}\), play a vital role in calculus due to their unique properties. The function \(e^x\) is the only function whose derivative and integral are both proportional to its value. This property makes it simpler to integrate.
When integrating exponential functions, the coefficient in the exponent often stays the same, simplifying calculations. This characteristic makes them easier to handle within integration techniques such as integration by parts.
Exponential decay (when the exponent is negative) is another important concept. The function \(e^{-x}\) illustrates this by showing how values decrease as \(x\) increases. It's commonly seen in real-life applications such as population modeling and radioactive decay.