8.1 Reversing differentiation
Integration is differentiation run backwards. In the last chapter, starting from \(y=ax^n\) we produced \(\frac {dy}{dx}=anx^{n-1}\). Here we are handed the derivative and asked to recover the function it came from.
Definition 8.1. If \(\frac {dy}{dx}=kx^n\), then \[y=\frac {kx^{n+1}}{n+1}+c,\qquad n\neq -1,\] where \(c\) is the constant of integration.
The rule says: add one to the index, then divide by the new index. It is the power rule read in reverse, and it is worth checking that it works — differentiating \(\frac {kx^{n+1}}{n+1}\) brings the \(n+1\) down to cancel the one underneath, leaving \(kx^{n}\).
Note 8.2. The constant \(c\) is not optional decoration. Differentiating destroys constants: \(x^3+7\), \(x^3-2\) and \(x^3\) all have derivative \(3x^2\). Running the process backwards therefore cannot tell which constant was there, and \(+c\) records that honestly. Every indefinite integral carries one.
Note 8.3. The condition \(n\neq -1\) is there because \(n+1\) sits in a denominator, and \(n=-1\) would make it zero. That single excluded case is handled separately, and its answer is a logarithm rather than a power.
Example 8.4. Find \(y\) in each case: (a) \(\frac {dy}{dx}=x^6\), (b) \(\frac {dy}{dx}=x^{-3}\), (c) \(\frac {dy}{dx}=4x^3-6x+5\).
Solution. (a) Here \(k=1\) and \(n=6\). Adding one to the index gives \(7\), and we divide by it: \[y=\frac {x^{7}}{7}+c.\]
(b) Here \(n=-3\), so \(n+1=-2\): \[y=\frac {x^{-2}}{-2}+c=-\frac {1}{2x^2}+c.\] Negative indices follow the same rule; the only care needed is with the signs. Adding one to \(-3\) gives \(-2\), not \(-4\).
(c) A sum is integrated term by term: \[y=\frac {4x^{4}}{4}-\frac {6x^{2}}{2}+5x+c=x^4-3x^2+5x+c.\] The term \(5\) is \(5x^0\), so the rule gives \(\frac {5x^1}{1}=5x\). Only one constant \(c\) is needed at the end, not one per term.
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