Problem 24 - Calculus for Biology and Medicine

Question

Use integration by parts to evaluate the integrals. $$ \int_{0}^{3} x^{2} e^{-x} d x $$

Step-by-Step Solution

Answer

The integral evaluates to approximately 1.1534.

Step 1: Choose Functions for Integration by Parts

Integration by parts is based on the formula \( \int u \, dv = uv - \int v \, du \). Here, we choose \( u = x^2 \) and \( dv = e^{-x} \, dx \).

Step 2: Differentiate and Integrate Components

Differentiate \( u = x^2 \) to get \( du = 2x \, dx \). Integrate \( dv = e^{-x} \, dx \) to obtain \( v = -e^{-x} \).

Step 3: Apply Integration by Parts Formula

Substitute into the integration by parts formula: \( \int_{0}^{3} x^2 e^{-x} \, dx = \left[-x^2 e^{-x}\right]_{0}^{3} + \int_{0}^{3} 2x e^{-x} \, dx \).

Step 4: Evaluate Boundary Terms

Substitute the bounds into the boundary term: \([-x^2 e^{-x}]_{0}^{3} = [-(3^2)(e^{-3}) - (0^2)(e^{0})] = -9e^{-3}\).

Step 5: Repeat Integration by Parts

Apply integration by parts again on \( \int_{0}^{3} 2x e^{-x} \, dx \) by setting \( u = 2x \) and \( dv = e^{-x} \, dx \). This gives \( du = 2 \, dx \) and \( v = -e^{-x} \).

Step 6: Substitute Second Integration by Parts Result

Using the integration by parts formula again: \( \int_{0}^{3} 2x e^{-x} \, dx = \left[ -2x e^{-x} \right]_0^3 + \int_{0}^{3} 2 e^{-x} \, dx \).

Step 7: Evaluate Second Boundary Terms

Calculate \( [-2x e^{-x}]_{0}^{3} = [-(2 imes 3)(e^{-3}) - (2 imes 0)(e^{0})] = -6e^{-3} \).

Step 8: Integrate Remaining Term

Calculate \( \int_{0}^{3} 2 e^{-x} \, dx \). The integral of \( e^{-x} \) is \( -e^{-x} \), so \( \int_{0}^{3} 2 e^{-x} \, dx = [-2e^{-x}]_0^3 = [-2e^{-3} + 2] \).

Step 9: Compile Final Result

Substitute all parts back: \( -9e^{-3} - 6e^{-3} + (-2e^{-3} + 2) \). This simplifies to \( -17e^{-3} + 2 \).

Step 10: Calculate Numerical Result

Substitute the approximate value of \( e^{-3} \approx 0.0498 \) to find \( -17 \times 0.0498 + 2 \approx 1.1534 \).

Key Concepts

Definite Integrals

Definite integrals allow us to calculate the net area under a curve bounded by two points on the x-axis. When working with definite integrals, the upper and lower bounds define the limits of integration.
For example, in the expression \( \int_{0}^{3} x^2 e^{-x} dx \), the limits 0 and 3 indicate that we want to find the area from \(x=0\) to \(x=3\) under the curve \(x^2 e^{-x}\). Understanding definite integrals is essential because they help in evaluating quantities where boundaries or limits are crucial, such as calculating total distances or accumulated quantities over time.
Several properties make definite integrals beneficial, such as:

• They provide exact values that can be approximated numerically.
• They are used in numerous applications like physics, engineering, and probability.
• They obey the fundamental theorem of calculus, connecting differentiation and integration.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent, often seen in the form \(f(x) = a^x\). When the base is \(e\), the natural exponential constant, the function becomes \(f(x) = e^x\).
In the exercise \( \int_{0}^{3} x^2 e^{-x} dx \), the component \(e^{-x}\) represents an exponential decay function. It is significant because its derivative is proportional to its value, which is a unique property of exponential functions.
These functions play a pivotal role in various fields due to their growth or decay behavior. For instance:

• Modeling population growth where populations increase at rates proportional to their size.
• Describing the decay of radioactive substances over time.
• Analyzing interest compounded continuously in finance.

Comprehending how exponential functions work is vital for mastering calculus and further mathematical disciplines.

Integration Techniques

Integration techniques are strategies used to find integrals, especially when direct integration is challenging. A critical technique in this exercise is integration by parts, an advanced method derived from the product rule of differentiation.
The formula used is \( \int u \, dv = uv - \int v \, du \). It's particularly useful when the integrand is a product of two functions that individually are tricky to integrate directly.
Here's how you typically choose components for integration by parts:

• **Identify \(u\) and \(dv\):** Choose \(u\) to be the part of the integrand that simplifies when differentiated and \(dv\) as the part that can be integrated easily. For example, choosing \(u=x^2\) and \(dv=e^{-x} \, dx\).
• **Differentiate and Integrate:** Differentiate \(u\) to find \(du\) and integrate \(dv\) to find \(v\).
• **Apply the formula:** Substitute into the formula to simplify the integral step by step.

This technique is invaluable, particularly for solving definite integrals involving products of polynomial and exponential or trigonometric functions. Mastering various integration techniques is crucial for tackling complex integrals effectively.