Problem 5
Question
Find each indefinite integral. \(\int \sqrt{u} d u\)
Step-by-Step Solution
Verified Answer
The indefinite integral is \( \frac{2}{3} u^{3/2} + C \).
1Step 1: Recognize the form of the integrand
The integrand is \( \sqrt{u} \), which can be rewritten using an exponent. Recognize that \( \sqrt{u} = u^{1/2} \). This helps us to apply the power rule for integration.
2Step 2: Apply the power rule for integration
The power rule for integration states that \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \) where \( n eq -1 \). Here, \( n = \frac{1}{2} \), so apply the rule: \[ \int u^{1/2} \, du = \frac{u^{1/2 + 1}}{1/2 + 1} + C = \frac{u^{3/2}}{3/2} + C \]
3Step 3: Simplify the expression
Simplify the expression \( \frac{u^{3/2}}{3/2} \) by multiplying by the reciprocal of the fraction: \[ \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2} \] Therefore, the integral becomes \( \frac{2}{3} u^{3/2} + C \).
Key Concepts
Power Rule for IntegrationIntegrandExponentiation in Integrals
Power Rule for Integration
When dealing with integrals, one of the most essential tools is the power rule for integration. This rule can make the task of finding indefinite integrals significantly easier. The rule states:
This rule simplifies many integral problems because it converts the process into a straightforward calculation, as seen with the example \( \int u^{1/2} \, du \) becoming \( \frac{u^{3/2}}{3/2} + C \) after applying the power rule.
- For any expression of the form \( u^n \), the integral is calculated as \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \) where \( n eq -1 \).
This rule simplifies many integral problems because it converts the process into a straightforward calculation, as seen with the example \( \int u^{1/2} \, du \) becoming \( \frac{u^{3/2}}{3/2} + C \) after applying the power rule.
Integrand
The integrand is a crucial term in integration. It's the function you want to integrate. In simple terms, it's the 'inside' part of the integral that you apply the integration process to. When you see your expression, it generally looks like this: \( \int f(x) dx \). Here, \( f(x) \) is the integrand.
For the exercise, the integrand is \( \sqrt{u} \). Understanding the nature of the integrand helps in selecting the right method or rule to apply, such as possible simplifications or rewriting, as done by expressing \( \sqrt{u} \) as \( u^{1/2} \).
Recognizing patterns and forms in the integrand, like polynomial forms that suit the power rule for integration, can streamline the entire process of integration.
For the exercise, the integrand is \( \sqrt{u} \). Understanding the nature of the integrand helps in selecting the right method or rule to apply, such as possible simplifications or rewriting, as done by expressing \( \sqrt{u} \) as \( u^{1/2} \).
Recognizing patterns and forms in the integrand, like polynomial forms that suit the power rule for integration, can streamline the entire process of integration.
Exponentiation in Integrals
Exponentiation in the realm of integrals refers to the powers within the functions being integrated. This can appear intimidating initially, especially with roots or fractional exponents. However, rewriting expressions with exponents is a strategy to simplify integration.
For instance, rewrite \( \sqrt{u} \) as \( u^{1/2} \). This step might seem modest but is crucial as it allows for easy application of rules like the power rule.
For instance, rewrite \( \sqrt{u} \) as \( u^{1/2} \). This step might seem modest but is crucial as it allows for easy application of rules like the power rule.
- This makes manipulation and understanding of the expressions simpler.
- It gives a clear path to apply integration techniques.
Other exercises in this chapter
Problem 5
For each demand function \(d(x)\) and demand level \(x\) find the consumers' surplus. $$ d(x)=350-0.09 x^{2}, \quad x=50 $$
View solution Problem 5
Find the average value of each function over the given interval. \(f(x)=\frac{1}{x^{2}}\) on [1,5]
View solution Problem 6
Find each indefinite integral. \(\int e^{0.02 x} d x\)
View solution Problem 6
For each demand function \(d(x)\) and demand level \(x\) find the consumers' surplus. $$ d(x)=840-0.06 x^{2}, \quad x=100 $$
View solution