2013 Paper 2 Question 13

(a)

Let $y = x/2$

$dy = dx/2$

Therefore,

$I = 2\,\int_0^{\infty} e^{-y^2}\,dy$

Now, given

$\int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt \pi$

we have, due to the even integrand

$\int_{0}^{\infty}e^{-x^2}dx = \frac{1}{2}\sqrt \pi$

Thus,

$I = 2 \left( \frac{1}{2} \sqrt \pi \right) = \sqrt \pi$

(b)

(i)

The key is to rewrite the integrand and use integration by parts,

$I_n = \int_{-\infty}^{\infty} x^{n-1}\,x\,e^{-x^2}dx$

Now let $u =x^{n-1}$ and $dv/dx=x\,e^{-x^2}$

$du = (n-1)x^{n-2}\,dx$

and

$v = -\frac{1}{2}e^{-x^2}$

$I_n = \left[ -\frac{1}{2}x^{n-1}e^{-x^2} \right]_{-\infty}^{\infty} + \frac{1}{2} \int_{-\infty}^{\infty}(n-1) x^{n-2}e^{-x^2}\,dx$

The exponential dominates the polynomial, so the first term is zero, and the final integral belongs to the family of given integrals, so

$I_n = \frac{n-1}{2} \, I_{n-2}$

NST First Year Maths