Problem 49 - Calculus for Biology and Medicine

Question

Use either substitution or integration by parts to evaluate each integral. $$ \int x e^{-2 x} d x $$

Step-by-Step Solution

Answer

The integral \( \int x e^{-2x} \, dx \) is \(-\frac{x}{2} e^{-2x} - \frac{1}{4} e^{-2x} + C \).

Step 1: Identify Parts for Integration by Parts

To solve the integral \( \int x e^{-2x} \, dx \), we use integration by parts. The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \]Choose \( u = x \) and \( dv = e^{-2x} \, dx \). This way, \( du = dx \), and we need to differentiate \( u \) to find \( du \).

Step 2: Differentiate and Integrate

Differentiate \( u = x \) to get \( du = dx \). Integrate \( dv = e^{-2x} \, dx \) to find \( v \). The integral of \( e^{-2x} \, dx \) is \(-\frac{1}{2} e^{-2x} \). Therefore, \( v = -\frac{1}{2} e^{-2x} \).

Step 3: Apply Integration by Parts Formula

Now substitute \( u, du, v, \) and \( dv \) into the integration by parts formula:\[ \int x e^{-2x} \, dx = uv - \int v \, du \]Substitute the values:\[ = x \left(-\frac{1}{2} e^{-2x}\right) - \int \left(-\frac{1}{2} e^{-2x}\right) dx \]

Step 4: Simplify and Integrate

Simplify the expression:\[ = -\frac{x}{2} e^{-2x} + \frac{1}{2} \int e^{-2x} \, dx \]Now, integrate \( \int e^{-2x} \, dx \):\[ = -\frac{x}{2} e^{-2x} + \frac{1}{2} \left(-\frac{1}{2} e^{-2x}\right) + C \]

Step 5: Final Expression

Combine the terms:\[ = -\frac{x}{2} e^{-2x} - \frac{1}{4} e^{-2x} + C \]This is the final expression for the integral. Remember that \( C \) is the constant of integration.

Key Concepts

Integral Calculus

Integral Calculus is one of the major branches of calculus. It focuses on finding the total size, area, or value accumulated by an entity. While differential calculus concerns itself with rates of change, integral calculus helps find out what is accumulated if the change is continuous.
The fundamental theorem of calculus links these two branches by showing how differentiation and integration are essentially inverse processes. This means that if you know the derivative, you can find the original function by integrating and vice versa.

• Integral: It represents the area under a curve or function.
• Definite vs. Indefinite Integrals: Definite integrals calculate the area with specific bounds. Indefinite integrals represent a general form of antiderivative functions plus a constant C.
• Real-life Application: Used to calculate quantities like areas, volumes and more from given rates.

For the specific problem of \( \int x e^{-2x} \, dx \), integral calculus helps us understand how the function accumulates area despite being a complex expression.

Techniques of Integration

Integration may seem daunting at first, but with diverse techniques, it becomes manageable. Among these, Integration by Parts stands out when dealing with products of functions. It simplifies integrals of the form \( \int u \, dv \) by transforming them using the formula \( \int u \, dv = uv - \int v \, du \). This approach requires:

• Identifying parts: Selecting which function should be \( u \) and which should be \( dv \).
• Differentiation and Integration: Finding \( du \) by differentiating \( u \), and obtaining \( v \) by integrating \( dv \).
• Substitution and Simplification: Applying the formula, simplifying, and integrating the remainder.

For \( \int x e^{-2x} \, dx \), integration by parts enabled us to break down the integral into more manageable pieces by choosing \( u = x \) and \( dv = e^{-2x} \, dx \). The success in solving the integral hinges on accurate choices for \( u \) and \( dv \). Always remember, practice makes perfect when mastering these techniques!

Integration Methods

Integration methods cover a wide array of strategies to tackle various integral problems. While substitution is another common technique, Integration by Parts is crucial when confronted with products of polynomial and exponential functions or logarithmic and inverse trigonometric functions.
Let's explore these methods a bit further in relation to our problem:

• Integration by Parts: Used when you have a product of functions. For \( \int x e^{-2x} \, dx \), choosing \( u = x \) makes it simpler after differentiating.
• Substitution: Suitable for integrals involving composite functions, aiming for a simpler replacement variable.
• Understanding Context: The method choice often depends on the function's nature. Recognizing whether substitution or integration by parts is more efficient comes with experience.

In the given example, utilising Integration by Parts was not arbitrarily chosen; it efficiently handled the polynomial-exponential product, demonstrating the utility of diverse methods in solving integral calculus problems.