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Integration

Question 6 of 11

Beginning Integration — The Reverse of Differentiation

Question

Given \(f'(t) = 2 - t + t^2 - t^3 + t^4\), find \(f(t)\)

Show solution Hide solution Fully worked — 3 steps
  1. Express \(f(t)\) as an integral

    where \(t\) is the variable, other than \(x\)

    \[ f(t) = \int\left(2 - t + t^2 - t^3 + t^4\right)dt \]
  2. Split the integral into a sum

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

    \[ = \int 2\,dt - \int t\,dt + \int t^2\,dt - \int t^3\,dt + \int t^4\,dt \]
  3. Apply the rule to find \(f(t)\)

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

    \[ f(t) = 2t - \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + \frac{t^5}{5} + c \]

    That is, \(f(t)\) is \(2t\), minus \(t\) squared over 2, plus \(t\) cubed over 3, minus \(t\) to the power 4 over 4, plus \(t\) to the power 5 over 5, plus the constant of integration \(c\).