Problem 79 - Calculus and its applications

Question

Solve each integral. Each can be found using rules developed in this section, but some algebra may be required. $$ \int \frac{(t+3)^{2}}{\sqrt{t}} d t $$

Step-by-Step Solution

Answer

The integral is \(\frac{2}{5}t^{\frac{5}{2}} + 4t^{\frac{3}{2}} + 18t^{\frac{1}{2}} + C\).

Step 1: Expand the Numerator

Start by expanding the square in the numerator. We have \[(t + 3)^2 = t^2 + 6t + 9.\] Thus, the original integral becomes \[\int \frac{t^2 + 6t + 9}{\sqrt{t}} \, dt.\]

Step 2: Separate the Integral

Next, separate the integral into three different integrals: \[\int \frac{t^2}{\sqrt{t}} \, dt + \int \frac{6t}{\sqrt{t}} \, dt + \int \frac{9}{\sqrt{t}} \, dt.\] This makes it easier to solve each part individually.

Step 3: Simplify Each Term's Integral

Simplify each fraction by combining the powers;1. For \(\frac{t^2}{\sqrt{t}}\), this becomes \(t^{2 - \frac{1}{2}} = t^{\frac{3}{2}}\).2. For \(\frac{6t}{\sqrt{t}}\), this simplifies to \(6t^{1 - \frac{1}{2}} = 6t^{\frac{1}{2}}\).3. For \(\frac{9}{\sqrt{t}}\), this becomes \(9t^{-\frac{1}{2}}\).

Step 4: Integrate Each Term

Now, integrate each term:1. The integral of \(t^{\frac{3}{2}}\) is \(\frac{2}{5}t^{\frac{5}{2}}\).2. The integral of \(6t^{\frac{1}{2}}\) is \(6 \times \frac{2}{3}t^{\frac{3}{2}} = 4t^{\frac{3}{2}}\).3. The integral of \(9t^{-\frac{1}{2}}\) is \(9 \times 2t^{\frac{1}{2}} = 18t^{\frac{1}{2}}\).

Step 5: Combine the Results

Combine the results of each integral to form the general solution:\[\frac{2}{5}t^{\frac{5}{2}} + 4t^{\frac{3}{2}} + 18t^{\frac{1}{2}} + C\]where \(C\) is the constant of integration.

Key Concepts

Definite Integrals

Definite integrals calculate the net area under a curve over a specific interval. Unlike indefinite integrals that provide a family of functions, definite integrals have two boundary limits. These are usually noted at the top and bottom of the integral sign. For example, in \[ \int_{a}^{b} f(x) \, dx, \]the function \(f(x)\) is integrated from \(x = a\) to \(x = b\).

• Evaluate the function: To solve a definite integral, find the antiderivative of the function and evaluate it at the upper and lower bounds.
• Calculate the difference: Subtract the value of the antiderivative at the lower bound from its value at the upper bound.

This gives you the numerical value representing the total area, considering any parts below the x-axis as negative contributions.
Definite integrals are highly useful in solving problems related to physics and engineering, such as finding displacement, area, and work done by a force.

Indefinite Integrals

Indefinite integrals, or antiderivatives, provide a general formula for a set of functions that, when differentiated, return back to the original function. Unlike definite integrals, they do not have boundaries and include an arbitrary constant,usually represented as \(C\). For example, the indefinite integral of a function \(f(x)\) is represented as:\[ \int f(x) \, dx = F(x) + C, \]where \(F(x)\) is the antiderivative of \(f(x)\).
This constant \(C\) arises because indefinite integration is the reverse of differentiation, and any constant differentiates to zero.

• Find the antiderivative: Use integration rules to find a function whose derivative is the original function.
• Add the constant: Always remember to include the constant of integration, denoting the family of possible solutions.

Indefinite integrals help in deriving general solutions to problems, especially in calculus where solutions to differential equations are often sought.

Power Rule for Integration

The power rule is an essential method for integrating functions of the form \(x^n\). When applying the power rule, we calculate the antiderivative by increasing the exponent by one and dividing by the new exponent. This is expressed mathematically as:\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \]where \(n eq -1\) because integration of \(x^{-1}\) involves the natural logarithm.

• Increase the exponent: Add one to the power of \(x\).
• Divide by the new exponent: Divide by this increased power.
• Add the constant \(C\): Do not forget to include the constant of integration.

This method is straightforward and works well for polynomial functions.
It is vital when dealing with integrals involving polynomial expressions, just like the example problem, where expressions were simplified to apply this rule efficiently. This allows for easy computation of the indefinite integral, turning complex expressions into understandable solutions.