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Magazine Chapter 5: Problem 67
Evaluate the definite integral. $$ \int_{1}^{2} x \sqrt{x-1} d x $$
Short Answer
Expert verified
The value of the definite integral is \( \frac{16}{15} \).
Step by step solution
01
Determine the Appropriate Integration Method
The given integral involves expressions of the form \(x \sqrt{x-1}\). Observing the expression we recognize it can be simplified by substitution where one of the variables becomes a simpler expression. Consider substitution \( u = x - 1 \).
02
Perform the Substitution
Using the substitution \( u = x - 1 \), differentiate both sides to find \( du = dx \). Moreover, the limits of integration change correspondingly: when \( x = 1 \), \( u = 0 \); when \( x = 2 \), \( u = 1 \). Substitute into the integral to get \( \int_{0}^{1} (u+1) \sqrt{u} \, du \).
03
Simplify and Split the Integral
Expand \((u+1)\sqrt{u} = u^{3/2} + u^{1/2}\). The integral becomes \(\int_{0}^{1} (u^{3/2} + u^{1/2}) \, du\). This can be separated into two separate integrals: \(\int_{0}^{1} u^{3/2} \, du + \int_{0}^{1} u^{1/2} \, du\).
04
Integrate Each Term
Apply the power rule for integration to each term. The power rule states \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). Integrate each term separately: \(\int u^{3/2} \, du = \frac{u^{5/2}}{5/2} = \frac{2}{5}u^{5/2}\) and \(\int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3}u^{3/2}\).
05
Evaluate the Definite Integral
Now recombine the integrals and evaluate from the limits 0 to 1. So, the expression is \[ \left. \frac{2}{5}u^{5/2} \right|_0^1 + \left. \frac{2}{3}u^{3/2} \right|_0^1 \]. Evaluate it: \( \frac{2}{5}(1^{5/2}) + \frac{2}{3}(1^{3/2}) - (\frac{2}{5}(0^{5/2}) + \frac{2}{3}(0^{3/2})) \). This simplifies to \( \frac{2}{5} + \frac{2}{3} = \frac{6}{15} + \frac{10}{15} = \frac{16}{15} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Substitution
Integration by substitution is a method used to simplify integrating complex expressions. It's akin to the reverse chain rule in differentiation. In our problem, we tackle the integral \( \int_{1}^{2} x \sqrt{x-1} \, dx \) by substitution.
- First, identify a part of the integrand that could be replaced by a simpler variable. In this case, substituting \( u = x - 1 \), reduces the square root component \( \sqrt{x-1} \) to \( \sqrt{u} \).
- Differentiating, we obtain \( du = dx \), simplifying the integration process.
- The substitution also alters the integration limits: when \( x = 1 \), \( u = 0 \); when \( x = 2 \), \( u = 1 \).
Power Rule for Integration
The power rule for integration helps integrate terms of the form \( u^n \). It states that \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \) for any real number \( n eq -1 \).
- Applying this rule to our expression \((u+1)\sqrt{u}\), expand it to \(u^{3/2} + u^{1/2}\).
- Each term is integrated separately using the power rule:
- For \( \int u^{3/2} \, du \), apply the rule to get \( \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2} \).
- Similarly, for \( \int u^{1/2} \, du \), apply the rule to get \( \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2} \).
Change of Limits in Definite Integrals
Changing the limits of integration is a crucial step once substitution is utilized. This ensures that the definite integral covers the correct interval.
- When using substitution such as \( u = x - 1 \), the limits for \( x \) automatically adjust with respect to \( u \).
- Initially, the limits were from 1 to 2. After substitution, they change to 0 to 1, corresponding to the values of \( u \) when \( x = 1 \) and \( x = 2 \), respectively.
Evaluating Integrals
Evaluating definite integrals involves finding the definite value of the integral by substituting the upper and lower limits.
- Using the antiderivatives obtained \( \frac{2}{5}u^{5/2} \) and \( \frac{2}{3}u^{3/2} \), evaluate them from 0 to 1.
- Substitute 1 in place of \( u \) in both expressions to get \( \frac{2}{5} \times 1^{5/2} + \frac{2}{3} \times 1^{3/2} \).
- For both expressions, substituting 0 results in zero, simplifying evaluation.
- After calculating, sum up the integrated parts: \( \frac{2}{5} + \frac{2}{3} = \frac{16}{15} \).
