STEP I 1997 question 6 solution

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STEP I 1997 question 6 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1997 question 6 solution

$\displaystyle x^4(1-x)^4 = x^8-4x^7+6x^6-4x^5+x^4$ . The easiest way of doing the next bit would probably be long division, but that will be painful to layout in TeX. So, let's compare coefficients...

Coeff of $\displaystyle x^8: a_6 = 1$

Coeff of $\displaystyle x^7: a_5 = -4$

Coeff of $\displaystyle x^6: a_6+a_4 = 6 \implies a_4 = 5$

Coeff of $\displaystyle x^5: a_5+a_3 = -4 \implies a_3 = 0$

Coeff of $\displaystyle x^4: a_4+a_2 = 1 \implies a_2 = -4$

Coeff of $\displaystyle x^3: a_3+a_1 = 0 \implies a_1 = 0$

Coeff of $\displaystyle x^2: a_2+a_0 = 0 \implies a_0 = 4$

Coeff of $\displaystyle x^1: a_1 = 0$ (Which we knew it should be, but it never hurts to check).

Coeff of $\displaystyle x^0: a_0+b = 0 \implies b=-4$

Using this in the integral we get:

$\displaystyle \int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx = \int_0^1 x^6-4x^5+5x^4-4x^2+4 - \frac{4}{1+x^2} dx$

RHS = $\displaystyle \frac 17 - \frac 46 + 1 - \frac 43 + 4 - \pi = \frac {22}{7} - \pi$ as required.

$\displaystyle \int_0^1 x^4(1-x)^4 dx = \int_0^1 x^8-4x^7+6x^6-4x^5+x^4 dx \\ = \frac 19 - \frac 12 + \frac 67 - \frac 46 + \frac 15 \\ = \frac{70 - 315 + 540 - 420 + 126}{630} = \frac{1}{630}$

But $\displaystyle 1+x^2 > 1 \text{ for } x > 0$ and so $\displaystyle 0 < \int_0^1 \frac{x^4(1-x)^4}{1+x^2} < \int_0^1 x^4(1-x)^4 = 1 / 630$ .

So since we know $\displaystyle \pi = \frac{22}{7} - \frac{x^4(1-x)^4}{1+x^2}$ we deduce $\displaystyle \frac {22}{7} > \pi > \frac{22}{7} - \frac{1}{630}$ .