Problem 60 - Calculus and its applications

Question

Occasionally, integration by parts yields an integral of the form \(\int u d v\) that is identical to the original integral. In some cases, we can then solve for \(\int u d v\) algebraically. For example, to find \(\int 2^{x} e^{x} d x,\) we let \(u=2^{x}\) and \(d v=e^{x},\) so \(d u=(\ln 2) 2^{x} d x\) and \(v=e^{x} .\) Using integration by parts, we have $$ \int 2^{x} e^{x} d x=2^{x} e^{x}-\ln 2 \int 2^{x} e^{x} d x $$ Note that \(\int 2^{x} e^{x} d x\) appears twice. Adding \(\ln 2 \int 2^{x} e^{x} d x\) to $$ \begin{aligned} \int 2^{x} e^{x} d x+\ln 2 \int 2^{x} e^{x} d x &=2^{x} e^{x} \\ (1+\ln 2) \int 2^{x} e^{x} d x &=2^{x} e^{x} \\ \int 2^{x} e^{x} d x &=\frac{2^{x} e^{x}}{1+\ln 2}+C \end{aligned} $$ Use this method to evaluate the integrals in Exercises \(59-62\) $$ \int 5^{x} e^{2 x} d x $$

Step-by-Step Solution

Answer

\(\int 5^x e^{2x} dx = \frac{5^x e^{2x}}{2 + \ln 5} + C\)

Step 1: Assigning Variables for Integration by Parts

Let's choose \( u = 5^x \) and \( dv = e^{2x}dx \). Then, we differentiate and integrate to find \( du \) and \( v \).We have \( du = (\ln 5) 5^x dx \) and integrating \( dv \), \( v = \frac{1}{2}e^{2x} \).

Step 2: Apply Integration by Parts Formula

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Applying it here:\[ \int 5^x e^{2x} dx = 5^x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \cdot (\ln 5) 5^x dx \]

Step 3: Simplify the Resulting Integral

Recognize that the remaining integral on the right is similar to the initial integral:\[ \int \frac{1}{2} e^{2x} \cdot (\ln 5) 5^x dx = \frac{\ln 5}{2} \int 5^x e^{2x} dx \]

Step 4: Set Up the Equation for the Original Integral

Substitute back to form an equation with the original integral:\[ \int 5^x e^{2x} dx = \frac{1}{2}5^x e^{2x} - \frac{\ln 5}{2} \int 5^x e^{2x} dx \]

Step 5: Solve Algebraic Equation for the Integral

Combine the terms involving the integral to isolate it:\[ \int 5^x e^{2x} dx + \frac{\ln 5}{2} \int 5^x e^{2x} dx = \frac{1}{2} 5^x e^{2x} \]\[ \left(1 + \frac{\ln 5}{2}\right) \int 5^x e^{2x} dx = \frac{1}{2} 5^x e^{2x} \]Solve for the integral:\[ \int 5^x e^{2x} dx = \frac{\frac{1}{2}5^x e^{2x}}{1 + \frac{\ln 5}{2}} + C \]

Step 6: Simplify the Integral Expression

Simplify the expression:\[ \int 5^x e^{2x} dx = \frac{5^x e^{2x}}{2 + \ln 5} + C \]

Key Concepts

Algebraic Manipulation

In calculus, algebraic manipulation plays a vital role in solving integrals, especially when using techniques like integration by parts. This involves rearranging equations to isolate desired variables or expressions. When facing an integral that reappears after applying integration by parts, we can solve for it algebraically. For instance, if the integral \( \int u \, dv \) leads to an equation where the original integral is reintroduced on both sides, you can use algebraic techniques to extract it.
By grouping terms and factorizing, the integral in question can be isolated. In our exercise, we managed the terms by adding or subtracting expressions from both sides of an equation.
Once isolated, the solution for the integral can often be presented in a simplified fraction form, which ultimately expresses the relationship between the variables involved. This manipulation requires a careful step-by-step process to ensure accuracy and precision in deriving the integral's value.

Definite Integrals

Definite integrals are used to calculate the area under a curve over a specific interval. Unlike indefinite integrals that include an arbitrary constant (denoted as \( C \)), definite integrals will yield a specific numerical value.
In the context of integration by parts, once you've applied all steps of introducing and calculating the integral, whether definite or indefinite needs to be determined by the interval boundaries.
If limits of integration are given, denote them at the ends of the integral symbol. For example, when dealing with \( \int_{a}^{b} f(x) \, dx \), confirm the calculation includes evaluating the antiderivative at these boundary points.
Notably, when solving by integration with such boundary conditions, the terms within the calculation must be assessed at both intervals then subtracted to obtain the definitive area between them, ensuring the role of those boundaries is clearly executed in the solution.

Exponential Functions

Exponential functions appear frequently in calculus and often involve the base \( e \), the natural exponential constant approximately equal to 2.718. They are instrumental due to their unique property of being their own derivative and integral up to a constant factor.
In the context of integration by parts or solving integrals featuring exponentials, recognizing terms like \( e^{x} \) or \( a^{x} \) is key, where \( a \) is another constant base, such as 5 in our exercise. In such functions, differentiate appropriately to maintain the correct format.
When differentiating or integrating these functions within an integral, note the derivative rules, such as that for \( e^{kx} \), where it remains an exponential while accumulating terms, particularly coefficients. This property of exponential functions simplifies integration operations since you maintain the original function form.
Thus, exponential functions maintain a seamless flow in solving calculus problems, including more intricate procedures like integration by parts, through these inherent simplification properties.