Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 1 / 5

Let

I = \int_{0}^{\infty} e^{- x^{2}} d x

. Positivity allows the squared integral to be written as a first-quadrant double integral. In polar coordinates the Jacobian determinant is

r

, so

I^{2} = \int_{0}^{π /2} \int_{0}^{\infty} e^{- r^{2}} r d r d θ = \frac{π}{4}, I = \frac{π}{2} .

x e^{- x^{2}}

vanishing at both endpoints,

\int_{0}^{\infty} x^{2} e^{- x^{2}} d x = \frac{1}{2} \int_{0}^{\infty} e^{- x^{2}} d x = \frac{π}{4} .

The substitution

u = x^{2}

also gives the two antiderivatives

\int x e^{- x^{2}} d x = - \frac{1}{2} e^{- x^{2}} + C, \int x^{3} e^{- x^{2}} d x = - \frac{1}{2} (x^{2} + 1) e^{- x^{2}} + C .

These values also follow from the Gaussian moment recurrence and can be checked by differentiation.

b

The upper boundary follows

y = x

until its intersection with

y = 1/ x

(1, 1)

, then follows the hyperbola to

(2, 1/2)

. The lower boundary is

y = 0

. Therefore the two vertical-slice descriptions are

0 \leq x \leq 1

0 \leq y \leq x

, and

1 \leq x \leq 2

0 \leq y \leq 1/ x

Integrating with respect to

y

I = \int_{0}^{1} x^{3} e^{- x^{2}} d x + \int_{1}^{2} x e^{- x^{2}} d x = [- \frac{1}{2} (x^{2} + 1) e^{- x^{2}}]_{0}^{1} + [- \frac{1}{2} e^{- x^{2}}]_{1}^{2} .

Using the antiderivatives from part (a),

I = \frac{1 - e ^{- 1} - e ^{- 4}}{2} .

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