91Ó°ÊÓ

Integration by substitution is a fundamental technique used in calculus to simplify the integration process. It's particularly useful when dealing with complex functions.

In our exercise, we encounter the expression \( \int \sqrt{3x + 1} \, dx \). Direct integration can be challenging, so we introduce a substitution to make it easier. We choose \( u = 3x + 1 \), a substitution that simplifies the expression within the square root.

• By substituting, we get \( du = 3dx \), leading us to solve for \( dx \) as \( dx = \frac{1}{3}du \).
• This change of variables transforms the integral into a more approachable form: \( \int \sqrt{u} \cdot \frac{1}{3} \, du \).
• The advantage is simplification, turning our integral into a basic power rule form: \( \int \frac{1}{3} u^{1/2} \, du \).

The integration then proceeds smoothly, yielding a simpler antiderivative. Substituting back for \( u \), we reconnect our answer to the original variable \( x \). This step-by-step transformation highlights the power of substitution to tackle otherwise complex integration scenarios.

Differential equations involve equations with derivatives and are fundamental in describing various phenomena.

The exercise presents a scenario where a function's derivative is explicitly given as another function: \( f'(x) = \sqrt{3x + 1} \).

• Understanding such differential equations bridges integral and differential calculus. Here, we are tasked with finding a function \( f \) whose derivative matches \( \sqrt{3x + 1} \).
• By seeking the antiderivative, or the integral, we reverse the differentiation process to find \( f \).
• This process highlights the interconnection between derivatives and integrals: finding a particular solution to the differential equation given specific conditions.

Ultimately, solving differential equations often boils down to integration. In practice, this connects abstract mathematics to tangible problems, providing a tool to resolve rates of change and initial values.

An initial value problem (IVP) consists of a differential equation along with an additional condition, known as the initial condition.

In our exercise, the function must pass through a specific point, here it is \( (0,1) \). This initial condition helps in determining the constant of integration \( C \), present in the indefinite integral.

• The equation \( f(x) = \frac{2}{9} (3x + 1)^{3/2} + C \) reflects a family of solutions due to \( C \).
• The initial condition \( f(0) = 1 \) allows us to solve for \( C \), differentiating a specific solution from an infinite set.
• Substituting \( x = 0 \) and \( f(x) = 1 \) into the equation yields the specific value of \( C \), giving us \( C = \frac{7}{9} \).

This step is crucial because it tailors a broad mathematical solution to fit within provided physical or geometrical constraints, ensuring relevance to real-world applications.