Definite integration of product of terms

Question

Finding $$\int^{6}_{0}x(x-1)(x-2)(x-3)(x-4)(x-6)dx$$

My progress so far

$$x(x-4)(x-6)(x-1)(x-2)(x-3)=\bigg(x^3-10x^2+24x\bigg)\bigg(x^3-6x^2+11x-6\bigg)$$

How can I find solution Help me

Answer 1

Well, you can certainly proceed as you are now, and continue expanding until you get the polynomial into standard form, then integrate. We could go about it in a bit of a different way, though.

Let $f(x)=x(x-1)(x-2)(x-3)(x-4)(x-6).$ Then we have $$\int_0^6f(x)dx=\int_0^3f(x)dx+\int_3^6f(x)dx,$$ and making the substitution $x\mapsto 6-x$ (so $3\mapsto 3,$ $6\mapsto 0,$ and $dx\mapsto-dx$) in the rightmost integral gets us

\begin{eqnarray}\int_0^6f(x)dx &=& \int_0^3f(x)dx-\int_3^0f(6-x)dx\\ &=& \int_0^3f(x)dx+\int_0^3f(6-x)dx\\ &=& \int_0^3\bigl(f(x)+f(6-x)\bigr)dx.\end{eqnarray}

Now, note that

\begin{eqnarray}f(6-x) &=& (6-x)(5-x)(4-x)(3-x)(2-x)(0-x)\\ &=& (-1)^6(x-6)(x-5)(x-4)(x-3)(x-2)(x-0)\\ &=& x(x-2)(x-3)(x-4)(x-5)(x-6),\end{eqnarray}

so that, remembering our difference of squares formula $(u+v)(u-v)=u^2-v^2,$ we can see that

\begin{eqnarray}f(x)+f(6-x) &=& x(x-2)(x-3)(x-4)(x-6)\bigl((x-1)+(x-5)\bigr)\\ &=& x(x-2)(x-3)(x-4)(x-6)(2x-6)\\ &=& 2x(x-2)(x-3)^2(x-4)(x-6)\\ &=& 2x(x-6)(x-2)(x-4)(x-3)^2\\ &=& 2\bigl((x-3)+3\bigr)\bigl((x-3)-3\bigr)\bigl((x-3)+1\bigr)\bigl((x-3)-1\bigr)(x-3)^2\\ &=& 2\left((x-3)^2-9\right)\left((x-3)^2-1\right)(x-3)^2\\ &=& 2\left((x-3)^4-10(x-3)^2+9\right)(x-3)^2\\ &=& 2\left((x-3)^6-10(x-3)^4+9(x-3)^2\right).\end{eqnarray}

Thus, we have $$\int_0^6f(x)dx=2\int_0^3\left((x-3)^6-10(x-3)^4+9(x-3)^2\right)dx,$$ and making the substitution $x\mapsto 3-x$ (so that $0\mapsto 3,$ $3\mapsto 0,$ and $x\mapsto-dx$) on the right-hand side, we get

\begin{eqnarray}f(6-x) &=& -2\int_3^0\left((-x)^6-10(-x)^4+9(-x)^2\right)dx\\ &=& -2\int_3^0\left(x^6-10x^4+9x^2\right)dx\\ &=& 2\int_0^3\left(x^6-10x^4+9x^2\right)dx,\end{eqnarray}

which is a much nicer integral to evaluate. Still, that's a lot of work to go through just to avoid a few more polynomial expansions.

Answer 2

Hint: Expanding the Integrand we get $$x^6-16 x^5+95 x^4-260 x^3+324 x^2-144 x$$

Answer 3

If you just perform all the multiplication you get an integral of a sum of powers. The integral of a sum is the sum of the itegrals of each term, which will be of the formg $c\int \! \mathrm{d}x \; x^{\alpha} = c\;\frac{x^{\alpha+1}}{\alpha+1}$