Chapter 7: Problem 10
In Problems 1-54, perform the indicated integrations. \(\int_{0}^{4} \frac{5}{\sqrt{2 t+1}} d t\)
Short Answer
Expert verified
The integral evaluates to 10.
Step by step solution
01
Identify the integral to be solved
We are asked to find the integral of \( \int_{0}^{4} \frac{5}{\sqrt{2t+1}} \, dt \). The definite integral is calculated from the lower limit 0 to the upper limit 4.
02
Recognize the standard integration formula
Notice that the integrand \( \frac{5}{\sqrt{2t+1}} \) can be approached using the formula for \( \int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u} + C \). Here, substitution is the key.
03
Perform substitution
Let \( u = 2t + 1 \). Then, \( \frac{du}{dt} = 2 \) or \( dt = \frac{du}{2} \). This substitution helps transform the integral into a standard form.
04
Change the limits of integration
When \( t = 0 \), \( u = 2(0) + 1 = 1 \). When \( t = 4 \), \( u = 2(4) + 1 = 9 \). So, the limits of integration change from 0 to 4 to 1 to 9.
05
Rewrite the integral in terms of \( u \)
Substitute into the integral: \( \int_{1}^{9} \frac{5}{\sqrt{u}} \cdot \frac{1}{2} \, du = \frac{5}{2} \int_{1}^{9} u^{-1/2} \, du \).
06
Integrate the function
Using the integration formula \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), where \( n = -1/2 \), calculate the integral: \( \frac{5}{2} \left[ \frac{u^{1/2}}{1/2} \right]_{1}^{9} = 5 \left[ \sqrt{u} \right]_{1}^{9} \).
07
Evaluate the definite integral
Evaluate the result: \( 5 \left[ \sqrt{9} - \sqrt{1} \right] = 5 \left[ 3 - 1 \right] = 5 \times 2 = 10 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
Definite integrals provide a means to calculate the accumulation of quantities, such as areas under curves, between specified limits. In the expression \( \int_{a}^{b} f(x) \, dx \), the lower limit \( a \) and the upper limit \( b \) define the interval over which the integral is evaluated. Unlike indefinite integrals, which include a constant of integration \( C \), definite integrals result in a specific numerical value. This value represents the total change or area, considering the limits.
- The process involves taking a function, \( f(x) \), and evaluating its accumulation from \( a \) to \( b \).
- Important: The limits of integration determine where this area ends and begins.
U-Substitution Method
The u-substitution method is a technique used to simplify the process of integration, making complex integrals more manageable. It involves replacing part of the original integral with a single variable, \( u \), leading to a simpler expression. Here's a breakdown of the method:
- Identify the substitution: Choose a substitution \( u \), that simplifies the integrand. For example, in \( \int \frac{5}{\sqrt{2t+1}} \, dt \), setting \( u = 2t + 1 \) helps reduce the expression.
- Differentiate: Compute \( \frac{du}{dt} \) and rearrange to solve for \( dt \).
- Transform the integral: Substitute \( u \) and \( dt \) into the original integral.
Integration Formulas
Integration relies on various formulas that serve as shortcuts for finding anti-derivatives (the reverse process of differentiation). These formulas are crucial in calculus as they provide direct solutions for common types of integrands.
- Basic Integral Formulas: Examples include \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for \( n eq -1 \), \( \int e^x \, dx = e^x + C \), etc.
- Special Integrals: The formula used in our problem, \( \int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u} + C \), shows how certain integrals map directly to simpler functions.
Limits of Integration
Limits of integration define the scope of analysis for a definite integral. These are the values at which you start and stop calculating the area or accumulation. In definite integrals, once we perform a substitution like in u-substitution, these limits must change according to the new variable.
- Determine New Limits: When we substitute \( u = 2t + 1 \), the limits change from \( t \)-space to \( u \)-space. So, original \( t=0 \) becomes \( u=1 \), and \( t=4 \) becomes \( u=9 \).
- Evaluate with New Limits: Solve the integral using these transformed limits.