Problem 8 - Calculus Single Variable

Question

Determine a substitution that will simplify the integral. In each problem, record your choice of \(u\) and the resulting expression for \(d u .\) Then evaluate the integral. $$ \int 24 \frac{\sqrt{t}+1}{\sqrt{t}} d t $$

Step-by-Step Solution

Answer

The evaluated integral is \(24t + 48\sqrt{t} + C\).

Step 1: Identify the Inner Function

We look at the integrand \(\frac{\sqrt{t}+1}{\sqrt{t}}\) to identify an expression within it that can simplify the integral if substituted. Notice that \(\sqrt{t} = t^{1/2}\). If we choose \(u = \sqrt{t} = t^{1/2}\), this will simplify the expression.

Step 2: Determine the Derived Term for Substitution

Since we have chosen \(u = t^{1/2}\), we need to find \(du\). Differentiating both sides of \(u = t^{1/2}\), we get \(du = \frac{1}{2}t^{-1/2}dt\), or equivalently, \(dt = 2t^{1/2}du = 2udu\).

Step 3: Substitute into the Integral

Substitute \(u = t^{1/2}\) and \(dt = 2u du\) into the integral: \[\int 24 \frac{\sqrt{t}+1}{\sqrt{t}} \, dt = \int 24 \frac{u+1}{u} \, 2u \, du\] This simplifies to: \[\int 48(u+1) \, du = \int 48u + 48 \, du\] which results in: \[48 \int u \, du + 48 \int \, du\]

Step 4: Evaluate the Simplified Integral

Now, solve each part of the integral: - \(48 \int u \, du = 48 \cdot \frac{u^2}{2} = 24u^2\)- \(48 \int \, du = 48u\)Putting it all together, we get:\[24u^2 + 48u + C\] where \(C\) is the constant of integration.

Step 5: Back-Substitute \(u\) to Original Variable

Since \(u = \sqrt{t}\), substitute back to express your result in terms of \(t\): \[24(\sqrt{t})^2 + 48(\sqrt{t}) + C = 24t + 48\sqrt{t} + C\]

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics focusing on integrals and their properties, and it forms one of the two fundamental areas of calculus, the other being differential calculus. In general, integral calculus is concerned with two main concepts:

• Finding the accumulation of quantities, such as areas under curves.
• Reversing the process of differentiation.

In simpler terms, integration allows us to measure areas under curves and solve problems related to accumulation, such as finding total distance from velocity or overall growth from a rate of change. There are several techniques for solving integrals, and choosing the right one depends on the function being integrated. These techniques include:

• Basic integration rules known as anti-differentiation.
• Methods like substitution and integration by parts when basic rules are insufficient.

By understanding integral calculus, students learn to solve complex problems involving areas and accumulations, turning difficult questions into manageable ones through the application of known principles and formulas.

U-Substitution

U-substitution is a powerful technique used in integral calculus to simplify the process of indefinite integrals. It is particularly useful when an integral involves a composite function. This method relies on substituting a part of the integrand with a single variable, typically denoted as \( u \), which simplifies the integral into a more familiar form.**Why Use U-Substitution?**
U-substitution helps in simplifying an integral, so it's easier to solve. When using substitution:

• Identify a part of the integrand to replace, usually an expression inside or involving a function.
• Express the chosen part with the variable \( u \).
• Differentiate this expression to find \( du \), helping to transform both parts of the integration variable.

In our original exercise, we chose \( u = \sqrt{t} \), which is the key to simplifying the integrand \( \int 24 \frac{\sqrt{t}+1}{\sqrt{t}} \, dt \). This substitution turns the integrand from a complex expression to a simpler polynomial form \( \int 48(u+1) \, du \), thus making the integration process straightforward.

Definite and Indefinite Integrals

In integral calculus, integrals can be classified into two categories: definite and indefinite integrals. Understanding these types is crucial for applying integration effectively in different contexts.**Indefinite Integrals**
An indefinite integral represents a family of functions and includes an arbitrary constant, \( C \). It is the reversed process of differentiation. For example, \( \int f(x) \, dx = F(x) + C \), where \( F(x) \) is an antiderivative of \( f(x) \).When solving the original exercise, the result \( 24t + 48\sqrt{t} + C \) is an indefinite integral because it includes the constant \( C \), implying a family of functions differing by a constant.**Definite Integrals**
Unlike indefinite integrals, definite integrals calculate a specific numerical value representing the area under a curve. Given limits \( a \) and \( b \), it is expressed as:\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]This represents the accumulation of the function from \( a \) to \( b \).Understanding the difference helps in solving different types of problems. Indefinite integrals are used for finding antiderivatives, while definite integrals are applied when an exact figure or area is needed within specified limits.