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Write the given definite integral.

Mathematics

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Answer

73−ln⁡2\frac{7}{3} - \ln 2

Step-by-step solution

4 steps

  1. Write the given definite integral.

    ∫x=1x=2(x2−1x)dx\int_{x=1}^{x=2} \left(x^2 - \frac{1}{x}\right) dx

  2. Find the antiderivative of each term in the integrand.

    The antiderivative of x2x^2 is x2+12+1=x33\frac{x^{2+1}}{2+1} = \frac{x^3}{3}.
    The antiderivative of 1x\frac{1}{x} is ln⁡∣x∣\ln|x|.
    So, the antiderivative of (x2−1x)\left(x^2 - \frac{1}{x}\right) is x33−ln⁡∣x∣\frac{x^3}{3} - \ln|x|.

  3. Evaluate the definite integral using the Fundamental Theorem of Calculus.

    [x33−ln⁡∣x∣]12\left[\frac{x^3}{3} - \ln|x|\right]_{1}^{2}

    Substitute the upper limit (x=2x=2) and subtract the result of substituting the lower limit (x=1x=1).

    (233−ln⁡∣2∣)−(133−ln⁡∣1∣)\left(\frac{2^3}{3} - \ln|2|\right) - \left(\frac{1^3}{3} - \ln|1|\right)

  4. Simplify the expression.

    (83−ln⁡2)−(13−0)\left(\frac{8}{3} - \ln 2\right) - \left(\frac{1}{3} - 0\right)

    Since ln⁡1=0\ln 1 = 0.

    83−ln⁡2−13\frac{8}{3} - \ln 2 - \frac{1}{3}

    Combine the fractional terms.

    (83−13)−ln⁡2\left(\frac{8}{3} - \frac{1}{3}\right) - \ln 2 73−ln⁡2\frac{7}{3} - \ln 2
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