Lesson Explainer: Indefinite Integrals: The Power Rule Mathematics • Higher Education

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 π‘ˆβŠ†β„, we call a function πΉβˆΆπ‘ˆβ†’β„ the antiderivative of 𝑓 if 𝐹(π‘₯)=𝑓(π‘₯) for every π‘₯βˆˆπ‘ˆ. It follows that for any constant C, the function 𝐺(π‘₯)=𝐹(π‘₯)+C is also an antiderivative of 𝑓. This is expressed by the expression 𝑓(π‘₯)π‘₯=𝐹(π‘₯)+dC and we say that the expression 𝐹(π‘₯)+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 β€œπ‘₯”.

For example, the formula 𝑓(π‘₯)=3π‘₯ defines a function on π‘ˆ=ℝ.

Because ddπ‘₯ο€Ήπ‘₯=2π‘₯, we have ddπ‘₯ο€Ύπ‘₯2=π‘₯ and ddπ‘₯ο€Ύ3π‘₯2=3π‘₯. In other words, 𝐹(π‘₯)=𝑓(π‘₯)=3π‘₯𝐹(π‘₯)=32π‘₯,when which is expressed in integral notation by 𝑓(π‘₯)π‘₯=𝐹(π‘₯)+ο„Έ3π‘₯π‘₯=32π‘₯+.dCordC

Recall the following properties of derivatives:

  1. ddddddπ‘₯(𝐹+𝐺)=π‘₯𝐹+π‘₯𝐺 for any functions 𝐹,𝐺.
  2. ddddπ‘₯(π‘ŽπΉ)=π‘Žπ‘₯𝐹 for any function 𝐹 and any constant π‘Ž.
  3. ddπ‘₯π‘₯=𝑛π‘₯ for 𝑛=1,2,…; in particular, ddπ‘₯(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): ο„Έ(𝑓(π‘₯)+𝑔(π‘₯))π‘₯=𝑓(π‘₯)π‘₯+𝑔(π‘₯)π‘₯ddd

andο„Έπ‘Žπ‘“(π‘₯)π‘₯=π‘Žο„Έπ‘“(π‘₯)π‘₯.dd

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): ο„Έπ‘₯π‘₯=π‘₯𝑛+1+,𝑛=0,1,….dCfor So, for example, ο„Έπ‘₯=π‘₯+dC because ο„Έπ‘₯=ο„Έ1π‘₯=ο„Έπ‘₯π‘₯=π‘₯0+1+=π‘₯+.dddCC

Every polynomial is just a sum of constant multiples of powers of π‘₯, so we combine these results in order to evaluate their indefinite integrals. For example: ο„Έο€Ή2π‘₯βˆ’π‘₯+4π‘₯=ο„Έο€Ή2π‘₯π‘₯+ο„Έ(βˆ’1)π‘₯π‘₯+ο„Έ(4)π‘₯()=2ο„Έπ‘₯π‘₯βˆ’ο„Έπ‘₯π‘₯+4ο„Έπ‘₯()=2ο€Ύπ‘₯3+1οŠβˆ’π‘₯1+1+4π‘₯+()=12π‘₯βˆ’12π‘₯+4π‘₯+.οŠͺddddadditivitydddscalarmultiplesCpowerfunctionsC

Example 1: Integrating Monomials

Determine ο„Έβˆ’π‘₯π‘₯d.

Answer

The polynomial is (βˆ’1)π‘₯. Its indefinite integral is, therefore, ο„Έβˆ’π‘₯π‘₯=(βˆ’1)ο€Ύπ‘₯9+1+=βˆ’π‘₯10+.dCC

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

Example 2: Integrating Polynomials

Determine ο„Έο€Ή25π‘₯βˆ’65π‘₯+36π‘₯d.

Answer

The integrand of the polynomial is 25π‘₯βˆ’65π‘₯+36. Its indefinite integral is, therefore, ο„Έ25π‘₯βˆ’65π‘₯+36π‘₯=ο„Έ2π‘₯π‘₯+ο„Έβˆ’65π‘₯π‘₯+ο„Έ36π‘₯=25ο„Έπ‘₯π‘₯+(βˆ’65)ο„Έπ‘₯π‘₯+36ο„Έπ‘₯=25ο€Ύπ‘₯3οŠβˆ’65ο€Ύπ‘₯2+36(π‘₯)+=253π‘₯βˆ’652π‘₯+36π‘₯+.dddddddCC

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

Example 3: Integrating Polynomials Involving Multiplying Out Brackets

Determine ο„Έ(π‘₯+4)ο€Ήπ‘₯βˆ’4π‘₯+16π‘₯d.

Answer

First, simplify the integrand by expanding the brackets: ο„Έ(π‘₯+4)ο€Ήπ‘₯βˆ’4π‘₯+16π‘₯=ο„Έο€Ήπ‘₯βˆ’4π‘₯+16π‘₯+4π‘₯βˆ’16π‘₯+64π‘₯=ο„Έο€Ήπ‘₯+64π‘₯.ddd

This integral can be performed directly: ο„Έο€Ήπ‘₯+64π‘₯=π‘₯4+64π‘₯+.οŠͺdC

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