Chapter 5: Problem 69
Evaluate. Each of the following can be integrated using the rules developed in this section, but some algebra may be required beforehand. $$ \int \frac{(t+3)^{2}}{\sqrt{t}} d t $$
Short Answer
Expert verified
\( \frac{2}{5}t^{5/2} + 4t^{3/2} + 18t^{1/2} + C \)
Step by step solution
01
Expand the numerator
First, expand \( (t+3)^2 \) using the binomial theorem. \[ (t+3)^2 = t^2 + 6t + 9 \]
02
Rewrite the integral
Rewrite the integral by substituting the expanded form of \( (t+3)^2 \): \[ \int \frac{t^2 + 6t + 9}{\sqrt{t}} \, dt \]
03
Simplify the integrand
Simplify each term in the numerator by dividing by \( \sqrt{t} \): \[ \int \left( \frac{t^2}{\sqrt{t}} + \frac{6t}{\sqrt{t}} + \frac{9}{\sqrt{t}} \right) \, dt = \int \left( t^{3/2} + 6t^{1/2} + 9t^{-1/2} \right) \, dt \]
04
Integrate each term separately
Integrate each term separately using the power rule: \[ \int t^{3/2} \, dt + 6 \int t^{1/2} \, dt + 9 \int t^{-1/2} \, dt \]
05
Apply the power rule
For each term, apply the power rule for integration \( \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \): \[ \int t^{3/2} \, dt = \frac{2}{5}t^{5/2}, \ \int t^{1/2} \, dt = \frac{2}{3}t^{3/2}, \ \int t^{-1/2} \, dt = 2t^{1/2} \]
06
Substitute back into the integral
Combine the integrals with the constants calculated in the previous step: \[ \int (t^{3/2} + 6t^{1/2} + 9t^{-1/2}) \, dt = \frac{2}{5}t^{5/2} + 6 \cdot \frac{2}{3}t^{3/2} + 9 \cdot 2t^{1/2} = \frac{2}{5}t^{5/2} + 4t^{3/2} + 18t^{1/2} + C \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
Understanding the definite integral is crucial in calculus. A definite integral represents the signed area under a curve between two specific points. It's written as \(\int_{a}^{b} f(x) \, dx\). The result is a number that indicates the net area between \(f(x)\) and the x-axis from \(x = a\) to \(x = b\). Key steps to solve a definite integral include:
- Understanding the function to be integrated.
- Determining the limits \(a\) and \(b\).
- Calculating the antiderivative \(F(x)\).
- Evaluating \(F(b) - F(a)\).
Indefinite Integral
Indefinite integrals are another vital part of calculus. They represent a family of functions and are written without specific limits. The general form is \(\int f(x) \, dx\). Instead of a numerical result, you get a function plus an arbitrary constant \(C\), since integration reverses differentiation and loses specific values. Let's break down the process:
- Identify the function to integrate.
- Find the antiderivative.
- Add the constant of integration \(C\).
Binomial Theorem
The binomial theorem is a powerful tool for expanding expressions raised to a power. It’s especially useful in algebra and calculus. The binomial theorem states:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \]
For example, to expand \( (t+3)^2 \) using the binomial theorem:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \]
- \(a\) and \(b\) are any numbers or variables.
- \(n\) is a non-negative integer.
- \(\binom{n}{k}\) is a binomial coefficient showing combinations.
For example, to expand \( (t+3)^2 \) using the binomial theorem:
- \(a = t \), \(b = 3 \), \(n = 2\).
- \(\binom{2}{0} t^2 + \binom{2}{1} t^1 3^1 + \binom{2}{2} 3^2\)
- Results in: \( t^2 + 6t + 9 \)
Power Rule
The power rule is a fundamental principle in differentiation and integration. To integrate using the power rule, you follow this simple formula:
\[ \int t^n \,dt = \frac{t^{n+1}}{n+1} + C \]
Here are the key points:
\[ \int t^n \,dt = \frac{t^{n+1}}{n+1} + C \]
Here are the key points:
- \(t\) is the variable.
- \(n\) is any real number except -1.
- A constant of integration \(C\) is added.
- Increase the exponent by 1: \(\frac{3}{2} + 1 = \frac{5}{2}\).
- Divide by the new exponent: \(\frac{t^{5/2}}{\frac{5}{2}} = \frac{2}{5}t^{5/2}\).