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Notely Maths 12
Class 12 Maths
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3.6 Q-8

Question Statement

Evaluate the integral: ∫23(x−1x)2,dx\int_{2}^{3} \left( x - \frac{1}{x} \right)^2 , dx

Background and Explanation

This problem involves integrating a squared expression. To simplify the integral, we will first expand the square to break it into manageable terms. Once expanded, we will use basic integration techniques to solve each term separately. Familiarity with the basic rules of integration, including the power rule and the integral of simple rational functions, is essential for solving this problem.

Solution

  1. Expand the Square:

    Begin by expanding the expression (x−1x)2\left( x - \frac{1}{x} \right)^2: (x−1x)2=x2−2x⋅1x+1x2\left( x - \frac{1}{x} \right)^2 = x^2 - 2x \cdot \frac{1}{x} + \frac{1}{x^2}

    Simplifying the middle term: =x2−2+1x2= x^2 - 2 + \frac{1}{x^2}

  2. Rewrite the Integral:

    Now, substitute the expanded expression into the integral: ∫23(x2−2+1x2),dx\int_{2}^{3} \left( x^2 - 2 + \frac{1}{x^2} \right) , dx

    Split this into three separate integrals: =∫23x2,dx−2∫231,dx+∫231x2,dx= \int_{2}^{3} x^2 , dx - 2 \int_{2}^{3} 1 , dx + \int_{2}^{3} \frac{1}{x^2} , dx

  3. Integrate Each Term:

    Now, we integrate each term separately:

    • For ∫23x2,dx\int_{2}^{3} x^2 , dx, use the power rule: ∫x2,dx=x33\int x^2 , dx = \frac{x^3}{3}

    • For ∫231,dx\int_{2}^{3} 1 , dx, the integral of 1 is simply xx: ∫1,dx=x\int 1 , dx = x

    • For ∫231x2,dx\int_{2}^{3} \frac{1}{x^2} , dx, recall that ∫x−2,dx=−x−1\int x^{-2} , dx = -x^{-1}: ∫1x2,dx=−1x\int \frac{1}{x^2} , dx = -\frac{1}{x}

  4. Substitute the Limits of Integration:

    Now, evaluate each term at the limits of 2 and 3:

    • For x33\frac{x^3}{3} from 2 to 3: 33−233=27−83=193\frac{3^3 - 2^3}{3} = \frac{27 - 8}{3} = \frac{19}{3}

    • For −2x-2x from 2 to 3: −2(3−2)=−2(1)=−2-2(3 - 2) = -2(1) = -2

    • For −1x-\frac{1}{x} from 2 to 3: −(13−12)=−(2−36)=16-\left( \frac{1}{3} - \frac{1}{2} \right) = -\left( \frac{2 - 3}{6} \right) = \frac{1}{6}

  5. Combine the Results:

    Now, combine the results of all the integrals: 193−2+16\frac{19}{3} - 2 + \frac{1}{6}

    To simplify, find a common denominator (6): 193=386,−2=−126,16=16\frac{19}{3} = \frac{38}{6}, \quad -2 = \frac{-12}{6}, \quad \frac{1}{6} = \frac{1}{6}

    So: 386−126+16=38−12+16=276=92\frac{38}{6} - \frac{12}{6} + \frac{1}{6} = \frac{38 - 12 + 1}{6} = \frac{27}{6} = \frac{9}{2}

    Therefore, the final result is: 92\boxed{\frac{9}{2}}

Key Formulas or Methods Used

  • Power Rule for Integration:
    ∫xn,dx=xn+1n+1\int x^n , dx = \frac{x^{n+1}}{n+1}

  • Basic Integrals:
    ∫1,dx=x,∫1x2,dx=−1x\int 1 , dx = x, \quad \int \frac{1}{x^2} , dx = -\frac{1}{x}

Summary of Steps

  1. Expand the squared term: (x−1x)2=x2−2+1x2\left( x - \frac{1}{x} \right)^2 = x^2 - 2 + \frac{1}{x^2}.
  2. Split the integral into three parts:
    ∫23x2,dx,∫231,dx,∫231x2,dx\int_{2}^{3} x^2 , dx, \quad \int_{2}^{3} 1 , dx, \quad \int_{2}^{3} \frac{1}{x^2} , dx
  3. Integrate each term separately.
  4. Substitute the limits of integration and simplify.
  5. Combine the results to get the final answer:
    92\boxed{\frac{9}{2}}