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Integrating Polynomials

Integrating Polynomials

Before getting into the specifics of how to integrate polynomials, it may be helpful to look at how polynomials are differentiated, as it is often easier to see integration as the inverse of differentiation. Let's differentiate \(x^n\) from first principles.

\[ \frac{\mathrm{d}}{\mathrm{d}x}x^n = \lim\limits_{\delta x \to 0} \frac{(x+\delta x)^n - x^n}{\delta x},\]

by the definition of differentiation from first principles. Let us expand

\[ (x+\delta x)^n \]

using a binomial expansion, as then we can deal termwise.

\[ (x+\delta x)^n = x^n + \binom{n}{1}x^{n-1}(\delta x) +\binom{n}{2}x^{n-2}(\delta x)^2 + \dots + (\delta x)^n. \]

Filling this in, we get

\[ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}x^n &= \lim\limits_{\delta x \to 0} \frac{(x+\delta x)^n - x^n}{\delta x} \\ &= \lim\limits_{\delta x \to 0} \frac{x^n + \binom{n}{1}x^{n-1}(\delta x) +\binom{n}{2}x^{n-2}(\delta x)^2 + \dots + (\delta x)^n - x^n}{\delta x} \\ &= \lim\limits_{\delta x \to 0} \frac{\binom{n}{1}x^{n-1}(\delta x) +\binom{n}{2}x^{n-2}(\delta x)^2 + \dots + (\delta x)^n}{\delta x} \\ &=\lim\limits_{\delta x \to 0} \left[ \binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-2}(\delta x) + \dots + (\delta x)^{n-1}\right]. \end{align}\]

Inspecting this termwise, all terms except the first contain \(\delta x\), so as we take the limit as \(\delta x\to 0\), we will be left with only the first term, ie.

\[\lim\limits_{\delta x \to 0} \left[ \binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-2}(\delta x) + \dots + (\delta x)^{n-1}\right] = \binom{n}{1}x^{n-1} .\]

We can easily evaluate

\[\begin{align} \binom{n}{1} &= \frac{n!}{(n-1)!\cdot 1} \\ &= n. \end{align}\]

From this, we can now arrive at a well-known formula, which is commonly known as 'the power rule'. This states

\[ \frac{\mathrm{d}}{\mathrm{d}x}x^n = nx^{n-1}.\]

How does this relate to integrating polynomials? We can integrate both sides with respect to \(x\), and we get

\[ \int \frac{\mathrm{d}}{\mathrm{d}x}x^n \, \mathrm{d}x = \int nx^{n-1} \, \mathrm{d}x .\]

On the left-hand side, the integral and derivative cancel each other out, giving

\[ x^n + C= \int nx^{n-1} \, \mathrm{d}x \]

(remembering to add the constant of integration). We want to find as a general formula the integral of \(x^n\), so let us substitute \(t = n-1\) (both \(t \) and \(n\) are dummy variables, so we can do this). This gives

\[ (t+1)\int x^t\, \mathrm{d}x = x^{t+1}+C.\]

Let us divide by \(t+1\) to get

\[ \int x^t\, \mathrm{d}x =\frac{1}{t+1} x^{t+1}+C.\]

(we rescale our constant \(C\) here). We can replace \(t\) by \(n\) here (as we will just be filling in numbers for \(t\) or \(n\)) to give the formula for integrating \(x^n\) as

\[ \int x^n\, \mathrm{d}x =\frac{1}{n+1} x^{n+1}+C.\]

Using the formula to integrate a polynomial

You may now be wondering how this applies to a general polynomial. Here is where we can use the property of linearity of integrals. In general, the linearity of a polynomial means we can use \(a\) and \(b\) constants and \(f\) and \(g\) functions to write:

\[ \int (af+bg)\, \mathrm{d}x = a\int f\, \mathrm{d}x +b \int g\, \mathrm{d}x .\]

You can apply this to a polynomial. A polynomial has the general form of

\[ a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 = \sum\limits_{i=0}^n a_ix^i ,\]


\[ \begin{align} & \int \left(a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 \right) \, \mathrm{d}x \\ &\quad = a_n\int x^n \, \mathrm{d}x + a_{n-1}\int x^{n-1}\, \mathrm{d}x + \dots + a_i\int x \, \mathrm{d}x + \int a_0 \, \mathrm{d}x . \end{align}\]

This means that

\[ \int \left( \sum\limits_{i=0}^n a_ix^i \right)\, \mathrm{d}x = \sum\limits_{i=0}^n a_i \int x^i \, \mathrm{d}x .\]

This formula may look complicated, but this all becomes far simpler in practice, as we will see through the following examples.

Examples of integrating polynomials

Integrate \(x^2\) with respect to \(x\).

We can fill this directly into our formula. We set \(n = 2\), which gives

\[ \begin{align} \int x^2 \, \mathrm{d}x &= \frac{1}{1+2}x^{1+2} + C \\ &= \frac{1}{3}x^3 + C.\end{align}\]


\[ \int 3x^4 + 8x \, \mathrm{d}x .\]

Using linearity of integrals, we can say that the given integral is equivalent to

\[ 3 \int x^4 \, \mathrm{d}x + 8 \int x \, \mathrm{d}x .\]

We are now in a form where we can use the above-derived formula twice, once with \(n = 4\) and another with \(n = 1\). Applying this will give

\[ 3\cdot \frac{1}{5}x^5 + 8\frac{1}{2}x^2 + C.\]

Simplifying further leaves us with:

\[ \frac{3}{5}x^5 + 4x^2 + C.\]


\[\int_2^4 (x-1)^2(x+1)\, \mathrm{d}x\]

by a) expanding brackets and then integrating b) using the substitution \(u = x - 1\).

You should obtain the same answer both ways.

Method (a): Let's expand this bracket first, using a little cheat. Let us write \(x + 1\) as \((x - 1) + 2\). This means that

\[ \begin{align} (x-1)^2(x+1)&= (x-1)^2\left( (x-1)+2\right) \\ &= (x-1)^3 + 2(x-1)^2.\end{align}\]We can now use the binomial theorem to expand these terms. This leaves us with the result of:

\[ \begin{align} (x-1)^3 + 2(x-1)^2 &= x^3 - 3x^2 + 3x-1 + 2(x^2-2x+1) \\ &= x^3-x^2-x+1. \end{align}\]Then we need to evaluate the integral of

\[\begin{align} \int_2^4 (x-1)^2(x+1)\, \mathrm{d}x &= \int_2^4 x^3-x^2-x+1 \, \mathrm{d}x \\ &= \int_2^4 x^3 \, \mathrm{d}x - \int_2^4 x^2 \, \mathrm{d}x - \int_2^4 x \, \mathrm{d}x + \int_2^4 1 \, \mathrm{d}x \\ &= \left.\frac{1}{4}x^4 \right|_{x=2}^{x=4} - \left.\frac{1}{3}x^3 \right|_{x=2}^{x=4} - \left.\frac{1}{2}x^2 \right|_{x=2}^{x=4} + \left.x \right|_{x=2}^{x=4} \\ &= (64-4) - \left(\frac{64}{3} - \frac{8}{3}\right) - (8-2) + (4-2) \\ &= 60 - \frac{56}{3} - 6 + 2 \\ &= \frac{112}{3}.\end{align} \]

Method (b): If \(u = x - 1\), then

\[ \frac{ \mathrm{d}u }{\mathrm{d}x } = 1,\]

then \(\mathrm{d}u = \mathrm{d}x \). We must also remember to switch the limits, going from \(2\) and \(4\) to \(1\) and \(3\). Using this substitution, we get

\[\int_2^4 (x-1)^2(x+1)\, \mathrm{d}x = \int_1^3 u^2(u+2)\, \mathrm{d}u. \]

We now expand and use the linearity of integrals to find

\[ \begin{align} \int_1^3 u^2(u+2)\, \mathrm{d}u &= \int_1^3 u^3 \, \mathrm{d}u + 2\int_1^3 u^2\, \mathrm{d}u \\ &= \left.\frac{1}{4}u^4 \right|_{u=1}^{u=3} + \frac{2}{3}\left.u^3 \right|_{u=1}^{u=3} \\ &= \frac{1}{4}(81-1) + \frac{2}{3}(27-1) \\ &= 20 + \frac{52}{3} \\ &= \frac{112}{3}. \end{align}\]

Both answers are the same, so we are satisfied that we have gone through the right steps both times.


\[ \int \frac{3}{5}x^{\frac{2}{3}} \, \mathrm{d} x.\]

\[ \begin{align} \int \frac{3}{5}x^{\frac{2}{3}} \, \mathrm{d} x &= \frac{3}{5}\int x^{\frac{2}{3}} \, \mathrm{d} x \\ &= \frac{3}{5} \cdot \frac{x^{\frac{2}{3}+1}}{\dfrac{2}{3}+1} + C \\ &= \frac{9}{25}x^{\frac{5}{3}} + C. \end{align}\]

Integrating Polynomials – Key takeaways

  • The formula for integrating \(x^n\) is given by \[ \int x^n\, \mathrm{d}x =\frac{1}{n+1} x^{n+1}+C.\]
  • Integration has the property of linearity, ie \[ \int (af+bg)\, \mathrm{d}x = a\int f\, \mathrm{d}x +b \int g\, \mathrm{d}x .\]
  • When applying linearity to a definite integral, remember to keep the limits on every integral.

Frequently Asked Questions about Integrating Polynomials

The formula for the integral of x^n still holds when n is fractional, so follow the same procedure.

Use the linearity of integrals to split the integral into integrals with a single term, and then use the formula  for integration of polynomials to evaluate each of these integrals.

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