Problem 24 - Calculus for Biology and Medicine

Question

In Problems 1-30, use integration by parts to evaluate each integral. $$ \int_{0}^{3} x^{2} e^{-x} d x $$

Step-by-Step Solution

Answer

The integral evaluates to approximately \( 2 - \frac{5}{e^{3}} \).

Step 1: Identify Functions for Integration by Parts

In integration by parts, we choose two functions based on the integral \( \int u \, dv = u v - \int v \, du \). For \( u \), we select a function that simplifies upon differentiation and for \( dv \), a function easy to integrate. Here, let \( u = x^2 \) and \( dv = e^{-x} \, dx \).

Step 2: Differentiate and Integrate

Differentiate \( u \): \( du = 2x \, dx \). Now integrate \( dv \): \( \int e^{-x} \, dx = -e^{-x} \), so \( v = -e^{-x} \).

Step 3: Apply the Integration by Parts formula

Plug \( u \), \( du \), \( v \), and \( dv \) into the formula: \[ \int u \, dv = u v - \int v \, du = x^2(-e^{-x}) - \int (-e^{-x})(2x) \, dx \].

Step 4: Simplify the Calculation

Compute \( x^2 (-e^{-x}) \) to get \( -x^2 e^{-x} \). Then, calculate the integral \( \int 2x e^{-x} \, dx \).

Step 5: Repeat Integration by Parts for Second Integral

For \( \int 2x e^{-x} \, dx \), set \( u = 2x \) and \( dv = e^{-x} \, dx \). Then \( du = 2 \, dx \) and \( v = -e^{-x} \). Repeat the process: \[ 2x(-e^{-x}) + \int 2(-e^{-x}) \, dx = -2x e^{-x} - 2 \int e^{-x} \, dx \].

Step 6: Evaluate Remaining Integral

For \( \int e^{-x} \, dx \), previously calculated as \(-e^{-x}\). Substitute back to find: \( -2x e^{-x} + 2e^{-x} \).

Step 7: Combine All Parts

Combine results: \[ -x^2 e^{-x} - (-2x e^{-x} + 2e^{-x}) \]. This simplifies to \(-x^2 e^{-x} + 2x e^{-x} - 2e^{-x} \).

Step 8: Evaluate Definite Integral from 0 to 3

Evaluate at bounds 3 and 0: \[ \left[-x^2 e^{-x} + 2x e^{-x} - 2e^{-x} \right]_0^3 \]. Calculate separately at 3 and 0: \((-(9)e^{-3} + 6e^{-3} - 2e^{-3}) - (0 + 0 - 2)\).

Step 9: Simplify and Arrive at Solution

Simplify calculations: \([-5e^{-3}] - [-2]\). That results in \( -5e^{-3} + 2 \). Numerical value using \( e \approx 2.71828 \) can be further simplified if required.

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus that help us determine the total accumulation of quantities. In simple terms, it calculates the area under a curve from one point to another. This is particularly useful in various fields, such as physics and engineering, where understanding the net accumulation or total change is crucial.

To calculate a definite integral, we need to evaluate the antiderivative at two bounds. In our exercise, the bounds are from 0 to 3. This process involves not only finding an indefinite integral but carefully substituting these boundaries to find the final result.

The definite integral formula is expressed as:

• \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \\]

Here, \( F(x) \) is the antiderivative of \( f(x) \). This formula tells us to plug in the upper bound \( b \) into \( F(x) \), subtract the value of \( F(x) \) at the lower bound \( a \). The result gives the net area between the curve \( f(x) \), the x-axis, and the vertical lines \( x = a \) and \( x = b \).

In our problem, this step was crucial to find the area under the curve \( x^{2} e^{-x} \) from 0 to 3.

Differentiation

Differentiation is another core principle of calculus, and it plays a significant role in evaluating integrals, especially when using integration by parts. Differentiation is essentially finding the rate at which a function changes at any point.

When using integration by parts, we start with a function that simplifies upon differentiation. In the step-by-step solution, differentiation was used to find \( du \) from our selected \( u = x^2 \). By differentiating, we determined that \( du = 2x \, dx \).

Understanding how to differentiate a function correctly is crucial when performing integration by parts, as it allows us to simplify complex integrals into manageable parts.

• The derivative of \( x^n \) is \( nx^{n-1} \). So, for \( x^2 \), it becomes \( 2x \).

Proper differentiation ensures that the integration process can be carried forward smoothly, transforming a challenging integral into an expressible solution.

Indefinite Integrals

An indefinite integral is essentially the reverse of differentiation. It represents a family of functions that can produce a given derivative. Indefinite integrals come without specified limits and always include an integration constant \( C \).

Within the integration by parts process, finding indefinite integrals is necessary. For instance, finding \( v \) involves integrating \( dv \). Starting from \( dv = e^{-x} \, dx \), we find that \( v = -e^{-x} \). This integral provides a function without specific bounds, which we later use within the bounds to find definite integrals.

Indefinite integrals are an essential step in the process, whereby:

• They are solved using basic integration rules, such as \( \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C \).

Thus, recognizing and calculating indefinite integrals correctly forms the backbone of higher calculus operations, merging seamlessly into methods like integration by parts.