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Chapter 4: Problem 40

Evaluate the integral. $$ \int_{1}^{5} \sqrt{2 x-1} d x $$

Short Answer

Expert verified
The value of the definite integral \(\int_{1}^{5} \sqrt{2 x-1} dx\) is \(\frac{26}{3}\).

Step by step solution

01

Apply substitution

Let's set: $$ u = 2x - 1 $$ Now differentiate u with respect to x: $$ \frac{du}{dx} = 2 $$ We can rewrite the differential dx as follows: $$ dx = \frac{du}{2} $$ Now, we also need to adjust our integral limits. When \(x = 1\), we have: $$ u = 2(1)-1 = 1 $$ And when \(x = 5\), we have: $$ u = 2(5)-1 = 9 $$ So the integral becomes: $$ \int_{1}^{5} \sqrt{2 x-1} dx = \frac{1}{2} \int_{1}^{9} \sqrt{u} du $$
02

Apply the power rule for integration

We can rewrite the integral as a power function: $$ \frac{1}{2} \int_{1}^{9} u^{1/2} du $$ Now we can apply the power rule for integration, which states that: $$ \int x^n dx = \frac{x^{n+1}}{n+1} + C $$ In our case, n is 1/2, therefore: $$ \frac{1}{2} \left[ \frac{u^{3/2}}{\frac{3}{2}} \right]_{1}^{9} $$ Simplify the expression and evaluate the integral at the limits: $$ \frac{1}{3} \left[ u^{3/2} \right]_{1}^{9} = \frac{1}{3} (9^{3/2} - 1^{3/2}) $$
03

Simplify and compute the final result

Now we can compute the result: $$ \frac{1}{3} (27 - 1) = \frac{1}{3} (26) = \frac{26}{3} $$ So the value of the definite integral is: $$ \int_{1}^{5} \sqrt{2 x-1} dx = \frac{26}{3} $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is a powerful technique used in calculus to simplify the process of integration, especially when dealing with complex functions. It is also often referred to as 'u-substitution'. In the original problem, substitution was used to transform the integral of \( \sqrt{2x-1} \) into a simpler form. Let's break it down:
  • First, identify a part of the integral to replace with a new variable. In this case, the expression \( 2x - 1 \) was substituted with \( u \).
  • Differentiate \( u \) to find \( \frac{du}{dx} = 2 \), which helps you solve for \( dx \), getting \( dx = \frac{du}{2} \).
  • The limits of the integral must also change based on the substitution. When \( x = 1 \), \( u = 1 \); and when \( x = 5 \), \( u = 9 \).
This technique simplifies the integral by transforming the original variables into something more manageable. After substitution, the integral becomes easier to solve, reducing the computational effort required.
Power Rule for Integration
The power rule for integration is an essential tool used when integrating expressions of the form \( x^n \). Often, after simplifying an integral using substitution, the expression is rewritten in a way that allows the power rule to be applied directly.
  • The general form of the power rule is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \).
  • In the given solution, the original integral \( \sqrt{u} \) was rewritten as \( u^{1/2} \) to apply the power rule.
  • Here, \( n = 1/2 \). Therefore, applying the rule gives \( \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} + C \).
When evaluating definite integrals, the constant \( C \) is not necessary, as the limits provide the specific value. The power rule simplifies the process of integrating powers, particularly useful after making substitutions.
Integral Limits
Integral limits refer to the boundaries at which the definite integral is evaluated. For definite integrals, these limits transform the process from finding indefinite integrals (which include a constant \( C \)) to computing an exact number.
  • Once substitution is made, the original limits need updating to match the new variable. This is crucial since keeping old limits would mean integrating along the wrong path.
  • Initially, the integral had limits \( 1 \) to \( 5 \), based on \( x \). When changing to \( u \), using \( u = 2x - 1 \), these transformed to \( u = 1 \) and \( u = 9 \).
  • Finally, to evaluate the definite integral \( \int_{1}^{9} \frac{u^{3/2}}{3/2} \, du \), compute \( F(9) - F(1) \) where \( F(u) \) is the antiderivative found from integration.
Adjusting the limits correctly ensures the result reflects the intended area under the curve described by the transformed function. It ties the process together, allowing the evaluation of the definite integral in terms of the new variable.

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