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Integration

Question 4 of 11

Beginning Integration — The Reverse of Differentiation

Question

Given \(f'(x) = \dfrac{1}{2}x^{\frac{1}{3}} - \dfrac{1}{4}x^{\frac{1}{5}} + \pi\), find \(f(x)\)

Show solution Hide solution Fully worked — 5 steps
  1. Express \(f(x)\) as an integral
    \[ f(x) = \int \left( \frac{1}{2}x^{\frac{1}{3}} - \frac{1}{4}x^{\frac{1}{5}} + \pi \right) dx \]
  2. Split the integral term by term

    Use \(\int f(x)\,dx + g(x)\,dx = \int f(x)\,dx + \int g(x)\,dx\):

    \[ = \int \frac{1}{2}x^{\frac{1}{3}}\,dx - \int \frac{1}{4}x^{\frac{1}{5}}\,dx + \int \pi\,dx \]
  3. Take the constants outside the integrals

    Use \(\int ax^{n}\,dx = a\int x^{n}\,dx = \frac{ax^{n+1}}{n+1} + c\):

    \[ = \frac{1}{2}\int x^{\frac{1}{3}}\,dx - \frac{1}{4}\int x^{\frac{1}{5}}\,dx + \pi\int dx \]
  4. Apply the rule to each term
    \[ = \frac{\frac{1}{2}x^{\frac{4}{3}}}{\frac{4}{3}} - \frac{\frac{1}{4}x^{\frac{6}{5}}}{\frac{6}{5}} + \frac{\pi x}{1} + c \]
  5. Simplify to find \(f(x)\)

    Divide \(\frac{1}{2}\) by \(\frac{4}{3}\) and \(\frac{1}{4}\) by \(\frac{6}{5}\):

    \[ f(x) = \frac{3x^{\frac{4}{3}}}{8} - \frac{5x^{\frac{6}{5}}}{24} + \pi x + c \]

    That is, \(f(x)\) is three \(x\) to the power four thirds over 8, minus five \(x\) to the power six fifths over 24, plus pi \(x\), plus the constant of integration \(c\).