# Mechanical Integration

To find the integral of a polynomial, evaluate it at some special
points and add multiples of these values.

There is a theory of polynomials which enables you to find the integral of a polynomial by simply evaluating it at some special points and adding certain multiples $\Lambda_1, \Lambda_2,\ldots$ of these values. For example, for all polynomials $q$ of degree less than six, the special points are $-\sqrt{3/5}$, $0$ and $+\sqrt{3/5}$, and $$\int_{-1}^1 q(x)dx = \Lambda_1q(-\sqrt{3/5}) + \Lambda_2q(0) + \Lambda_3q(+\sqrt{3/5}).\quad (1)$$ Find the multiples $\Lambda_1,\ \Lambda_2$ and $\Lambda_3$ by considering the three polynomials $q(x) = 1$, $q(x) = x$ and $q(x) = x^2$.

With these values of $\Lambda_1,\ \Lambda_2$ and $\Lambda_3$ show that the mechanical integration given by equation (1), which uses the values of the polynomial at the three special points, gives the value of the integral of ALL quadratic polynomials.

Now go on to show that the same formula gives the integral of ALL cubic, quartic and quintic polynomials.

Does the formula (1) hold for $q(x)=x^6$?

Taking $q(x)=1$, $q(x)=x$ and $q(x)=x^2$ in equation (1) and
working out the integral (easy!) will give you three equations
which you can solve to find $\lambda_1,\ \lambda_2$ and
$\lambda_3$.

The key to showing that the same formula works for other polynomials is to show that if it works for $1,\ x$ and $x^2$ it works for any linear combination of them and so for any quadratic polynomial.

Finally you can go on to check out the formula for $x^3,\ x^4$ and $x^5$.

The key to showing that the same formula works for other polynomials is to show that if it works for $1,\ x$ and $x^2$ it works for any linear combination of them and so for any quadratic polynomial.

Finally you can go on to check out the formula for $x^3,\ x^4$ and $x^5$.

John Lesieutre from State College Area High School, Pennsylvania, USA and Marcos Charalambides from Cyprus sent in excellent solutions to this problem.

Using $q(x)=1$ we get $2=\Lambda_1+\Lambda_2+\Lambda_3$.

Using $q(x)=x$ we get $0=\Lambda_1\sqrt{\frac{3}{5}}+\Lambda_3\sqrt{\frac{3}{5}}$ so $\Lambda_1=\Lambda_3$.

Using $q(x)=x^2$ we get $\frac{2}{3}=\frac{3\Lambda_1}{5}+\frac{3\Lambda_3}{5}= \frac{6\Lambda_1}{5}$ so $\Lambda_1=\frac{5}{9}=\Lambda_3$ and $\Lambda_2=\frac {8}{9}$.

Now if $q(x)=a+b x+c x^2$ (that is, a general quadratic), then $\int_{-1}^1 a+b x+c x^2 \mathrm{d}x=2a+\frac{2c}{3}$, and $\Lambda_1 q\left(-\sqrt{\frac{3}{5}}\right)+\Lambda_2 q\left(0\right)+\Lambda_3 q\left(\sqrt{\frac{3}{5}}\right)=\frac{5}{9}\left(a-b\sqrt{\frac{3}{5}}+\frac{3 c}{5}\right)+\frac{8a}{9}+\frac{5}{9}\left(a+b\sqrt{\frac{3}{5}}+\frac{3c}{5}\right) =\frac{10a}{9}+\frac{8a}{9}+\frac{2c}{3} =2a+\frac{2c}{3} $ so the formula works for all quadratics.

Using the same idea as above, to check that it works for cubics, quartics and quintics, we only have to check it for $x^3$, $x^4$ and $x^5$.

$\int_{-1}^1 x^3\mathrm{d} x = 0$ and $\frac{5}{9}\left(-\frac{3}{5}\sqrt{\frac {3}{5}}\right)+\frac{5}{9}\left(\frac{3}{5}\sqrt{\frac{3}{5}}\right)=0$.

$\int_{-1}^1 x^4\mathrm{d} x=\frac{2}{5}$ and $\frac{5}{9}\left(\frac{9}{25}\right) +\frac{5}{9}\left(\frac{9}{25}\right)=\frac{2}{5}$.

$\int_{-1}^1 x^5\mathrm{d} x=0$ and $\frac{5}{9}\left(-\frac{9}{25}\sqrt{\frac{ 3}{5}}\right)+\frac{5}{9}\left(-\frac{9}{25}\sqrt{\frac{3}{5}}\right)=0$.

$\int_{-1}^1 x^6\mathrm{d} x=\frac{2}{7}$ but $\frac{5}{9}\left(\frac{27}{125} \right)+\frac{5}{9}\left(\frac{27}{125}\right)=\frac{6}{25}$.

So the formula works for cubics, quartics and quintics, but not for higher powers.