Explainer: Indefinite Integrals: The Power Rule

In this explainer, we will learn how to find the indefinite integrals of polynomials and general power functions using the power rule for integration.

Recall that for any function 𝑓 defined on a subset 𝑈, we call a function 𝐹𝑈 the antiderivative of 𝑓 if 𝐹(𝑥)=𝑓(𝑥) for every 𝑥𝑈. It follows that for any constant C, the function 𝐺(𝑥)=𝐹(𝑥)+C is also an antiderivative of 𝑓. This is expressed by the expression 𝑓(𝑥)𝑥=𝐹(𝑥)+dC and we say that the expression 𝐹(𝑥)+C is the indefinite integral of 𝑓. This shorthand is particularly useful if the function 𝑓 is given by a formula involving an independent variable “𝑥”.

For example, the formula 𝑓(𝑥)=3𝑥 defines a function on 𝑈=.

Because dd𝑥𝑥=2𝑥, we have dd𝑥𝑥2=𝑥 and dd𝑥3𝑥2=3𝑥. In other words, 𝐹(𝑥)=𝑓(𝑥)=3𝑥𝐹(𝑥)=32𝑥,when which is expressed in integral notation by 𝑓(𝑥)𝑥=𝐹(𝑥)+3𝑥𝑥=32𝑥+.dCordC

Recall the following properties of derivatives:

  1. dddddd𝑥(𝐹+𝐺)=𝑥𝐹+𝑥𝐺 for any functions 𝐹,𝐺.
  2. dddd𝑥(𝑎𝐹)=𝑎𝑥𝐹 for any function 𝐹 and any constant 𝑎.
  3. dd𝑥𝑥=𝑛𝑥 for 𝑛=1,2,; in particular, dd𝑥(1)=0.

Properties (1) and (2) mean that the sum of antiderivatives is the antiderivative of a sum of functions (additivity), and that the antiderivative of a constant multiple of a function is that same constant multiplied by the antiderivative (scalar multiples): (𝑓(𝑥)+𝑔(𝑥))𝑥=𝑓(𝑥)𝑥+𝑔(𝑥)𝑥ddd


Of course, we must remember to add the constant C to every indefinite integral.

Property (3) is especially useful when the functions are polynomials. It leads us to the formula (power functions): 𝑥𝑥=𝑥𝑛+1+,𝑛=0,1,.dCfor So, for example, 𝑥=𝑥+dC because 𝑥=1𝑥=𝑥𝑥=𝑥0+1+=𝑥+.dddCC

Every polynomial is just a sum of constant multiples of powers of 𝑥, so we combine these results in order to evaluate their indefinite integrals. For example: 2𝑥𝑥+4𝑥=2𝑥𝑥+(1)𝑥𝑥+(4)𝑥()=2𝑥𝑥𝑥𝑥+4𝑥()=2𝑥3+1𝑥1+1+4𝑥+()=12𝑥12𝑥+4𝑥+.ddddadditivitydddscalarmultiplesCpowerfunctionsC

Example 1: Integrating Monomials

Determine 𝑥𝑥d.


The polynomial is (1)𝑥. Its indefinite integral is, therefore, 𝑥𝑥=(1)𝑥9+1+=𝑥10+.dCC

Notice how we only consider the constant C at the end.

Example 2: Integrating Polynomials

Determine 25𝑥65𝑥+36𝑥d.


The integrand of the polynomial is 25𝑥65𝑥+36. Its indefinite integral is, therefore, 25𝑥65𝑥+36𝑥=2𝑥𝑥+65𝑥𝑥+36𝑥=25𝑥𝑥+(65)𝑥𝑥+36𝑥=25𝑥365𝑥2+36(𝑥)+=253𝑥652𝑥+36𝑥+.dddddddCC

If the polynomial is not written in the standard form (as a sum of powers of 𝑥), it is necessary to do this first before evaluating the integral by this method.

Example 3: Integrating Polynomials Involving Multiplying Out Brackets

Determine (𝑥+4)𝑥4𝑥+16𝑥d.


First, simplify the integrand by expanding the brackets: (𝑥+4)𝑥4𝑥+16𝑥=𝑥4𝑥+16𝑥+4𝑥16𝑥+64𝑥=𝑥+64𝑥.ddd

This integral can be performed directly: 𝑥+64𝑥=𝑥4+64𝑥+.dC

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