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

Table of contents
- Solution b

Solution

The upper boundary follows

y = x

until its intersection with

y = 1/ x

at

(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

.

Figure 1.
The integration region below a line and a hyperbola
. Original shaded region with all four boundary pieces shown. Vertical slices change their upper endpoint at $(1, 1)$ . Details.

Integrating with respect to

y

first reduces the double integral to

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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