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Magazine Chapter 9: Problem 10
Evaluate the integral. $$ \int_{0}^{4} \frac{2 x+3}{\sqrt{1+2 x}} d x $$
Short Answer
Expert verified
The value of the integral is \( \frac{38}{3} \).
Step by step solution
01
Identify Integration Strategy
Let's look at the integral \( \int_{0}^{4} \frac{2x + 3}{\sqrt{1 + 2x}} \, dx \). The integral contains a rational function with a square root in the denominator. This suggests a substitution may simplify the expression. We will use substitution where \( u = 1 + 2x \).
02
Substitute and Simplify
Perform substitution where \( u = 1 + 2x \). Then, \( du = 2 \, dx \). Therefore, \( dx=\frac{1}{2} \, du \). The limits of integration change accordingly: when \( x = 0 \), \( u = 1 \); when \( x = 4 \), \( u = 9 \). Substitute these into the integral:\[ \int_{1}^{9} \frac{2x+3}{\sqrt{u}} \cdot \frac{1}{2} \, du \].
03
Express in Terms of u
Since \( u = 1 + 2x \), solve for \( x \) to find \( x = \frac{u - 1}{2} \). Substitute into the integral to express \( x \) in terms of \( u \). Thus:\[ 2x + 3 = 2\left(\frac{u - 1}{2}\right) + 3 = u - 1 + 3 = u + 2 \].Replace these back in the integrand:\[ \int_{1}^{9} \frac{u + 2}{\sqrt{u}} \cdot \frac{1}{2} \, du \].
04
Separate the Integral
Separate the integral into two simpler integrals:\[ \frac{1}{2} \int_{1}^{9} \frac{u}{\sqrt{u}} \, du + \frac{1}{2} \int_{1}^{9} \frac{2}{\sqrt{u}} \, du \]Simplify each term:1. \( \frac{u}{\sqrt{u}} = \sqrt{u} \)2. \( \frac{2}{\sqrt{u}} = 2u^{-1/2} \)
05
Integrate each part
Evaluate the two integrals separately:1. Integral of \( \sqrt{u} \): \[ \frac{1}{2} \int \sqrt{u} \, du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} = \frac{1}{3} u^{3/2} \].2. Integral of \( 2u^{-1/2} \): \[ \frac{1}{2} \int 2u^{-1/2} \, du = \int u^{-1/2} \, du = 2u^{1/2} \].
06
Evaluate Definite Integrals
Compute the values from \( u = 1 \) to \( u = 9 \):1. \( \frac{1}{3} \left[ u^{3/2} \right]_{1}^{9} = \frac{1}{3} ((9)^{3/2} - (1)^{3/2}) = \frac{1}{3} (27 - 1) = \frac{1}{3} \cdot 26 = \frac{26}{3} \).2. \( 2 \left[ u^{1/2} \right]_{1}^{9} = 2 ((9)^{1/2} - (1)^{1/2}) = 2(3-1) = 4 \).Add results: \( \frac{26}{3} + 4 = \frac{26}{3} + \frac{12}{3} = \frac{38}{3} \).
07
Final Result
Combine the computed results to find the value of the definite integral is \( \frac{38}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
A definite integral allows us to calculate the net area under a curve from one point to another. In this context, it involves integrating a function over a specific interval, in our case, from 0 to 4. By doing so, we obtain a number that represents this area, which may be positive if above the x-axis, negative if below, or a combination if the areas cancel each other out.
The expression of a definite integral involves three main parts:
The expression of a definite integral involves three main parts:
- the integral sign \( \int \)
- a function \( f(x) \)
- and two endpoints \( a \) and \( b \), forming \( \int_{a}^{b} f(x) \, dx \)
Substitution Method
The substitution method is a technique used to simplify the process of integration. This method changes the variable of integration to make the integral more approachable, often transforming it into a simpler form that is easier to evaluate.
In this exercise, we used the substitution \( u = 1 + 2x \). This effectively replaced the more complex terms in the original function with a simpler 'u', allowing us to recalibrate the bounds and express the integral in terms of \( u \).
Here鈥檚 how we use substitution:
In this exercise, we used the substitution \( u = 1 + 2x \). This effectively replaced the more complex terms in the original function with a simpler 'u', allowing us to recalibrate the bounds and express the integral in terms of \( u \).
Here鈥檚 how we use substitution:
- Identify the part of the integral that complicates it.
- Set \( u \) to an expression that includes this part (e.g., \( u = 1 + 2x \)).
- Find the derivative \( du \) to replace \( dx \) accordingly.
- Adjust the limits of integration in terms of \( u \).
Rational Functions
Rational functions are fractions of polynomials, such as \( \frac{2x + 3}{\sqrt{1 + 2x}} \). These types of functions frequently appear in calculus problems and often require techniques like substitution to integrate.
The integral in this problem has a rational function with a polynomial in the numerator and a square root of a polynomial in the denominator. These features make the function more complex and thus call for careful manipulation.
When dealing with rational functions:
The integral in this problem has a rational function with a polynomial in the numerator and a square root of a polynomial in the denominator. These features make the function more complex and thus call for careful manipulation.
When dealing with rational functions:
- Look for opportunities to simplify 鈥� decide if substitution or partial fraction decomposition is appropriate.
- Focus on the polynomial nature of the function to utilize algebraic techniques for simplification.
Square Root Functions
A square root function involves a function of the form \( \sqrt{something} \). In calculus, these often pose a challenge because they are non-linear and can complicate the integration process.
In our exercise, the square root appears in the form \( \sqrt{1 + 2x} \). It was one of the main reasons we chose substitution; by setting \( u = 1 + 2x \), we simplified \( \sqrt{1 + 2x} \) to \( \sqrt{u} \), making the integral more tractable.
Key strategies include:
In our exercise, the square root appears in the form \( \sqrt{1 + 2x} \). It was one of the main reasons we chose substitution; by setting \( u = 1 + 2x \), we simplified \( \sqrt{1 + 2x} \) to \( \sqrt{u} \), making the integral more tractable.
Key strategies include:
- Using substitution to eliminate the square root when possible.
- Rewriting the square root in terms of other expressions (e.g., power rules) that are easier to integrate.
