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Rewrite the integrand by dividing each term in the numerator by the denominator.

Mathematics

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Answer

274\frac{27}{4}

Step-by-step solution

4 steps

Here's the solution for question 2:

  1. Rewrite the integrand by dividing each term in the numerator by the denominator.

    ∫24(x3+3x2)dx=∫24(x3x2+3x2)dx\int_{2}^{4} \left(\frac{x^3+3}{x^2}\right)dx = \int_{2}^{4} \left(\frac{x^3}{x^2} + \frac{3}{x^2}\right)dx =∫24(x+3x−2)dx= \int_{2}^{4} \left(x + 3x^{-2}\right)dx

  2. Integrate each term with respect to xx.

    ∫(x+3x−2)dx=x1+11+1+3x−2+1−2+1+C\int \left(x + 3x^{-2}\right)dx = \frac{x^{1+1}}{1+1} + 3\frac{x^{-2+1}}{-2+1} + C =x22+3x−1−1+C= \frac{x^2}{2} + 3\frac{x^{-1}}{-1} + C =x22−3x+C= \frac{x^2}{2} - \frac{3}{x} + C

  3. Evaluate the definite integral using the limits of integration from 2 to 4.

    [x22−3x]24=(422−34)−(222−32)\left[ \frac{x^2}{2} - \frac{3}{x} \right]_{2}^{4} = \left(\frac{4^2}{2} - \frac{3}{4}\right) - \left(\frac{2^2}{2} - \frac{3}{2}\right)

  4. Simplify the expression.

    =(162−34)−(42−32)= \left(\frac{16}{2} - \frac{3}{4}\right) - \left(\frac{4}{2} - \frac{3}{2}\right) =(8−34)−(2−32)= \left(8 - \frac{3}{4}\right) - \left(2 - \frac{3}{2}\right) =(324−34)−(42−32)= \left(\frac{32}{4} - \frac{3}{4}\right) - \left(\frac{4}{2} - \frac{3}{2}\right) =294−12= \frac{29}{4} - \frac{1}{2} =294−24= \frac{29}{4} - \frac{2}{4} =274= \frac{27}{4}

    The value of the integral is 274\frac{27}{4}.

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