Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 15 / b

Here

y \geq 0

, so the absolute value equals

y

J = \int_{0}^{1} \int_{0}^{x} \int_{0}^{x y} x y d z d y d x .

The integrand is constant with respect to

z

J = \int_{0}^{1} \int_{0}^{x} x^{2} y^{2} d y d x = \frac{1}{3} \int_{0}^{1} x^{5} d x = \frac{1}{18} .

ii

Integrating first over

z

multiplies by

e^{y}

. Split at

y = 0

J = \int_{0}^{1} x [\int_{- x /2}^{0} (- y) e^{y} d y + \int_{0}^{x} y e^{y} d y] d x .

(1 - y) e^{y}

and

(y - 1) e^{y}

for these two inner integrands. Their contributions are

1 - (1 + x /2) e^{- x /2}, (x - 1) e^{x} + 1,

respectively. Thus

J = \int_{0}^{1} [2 x + (x^{2} - x) e^{x} - (x + x^{2} /2) e^{- x /2}] d x .

For the remaining integrals,

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

and

\int_{0}^{1} (x + x^{2} /2) e^{- x /2} d x = [- e^{- x /2} (x^{2} + 6 x + 12)]_{0}^{1} = 12 - 19 e^{- 1/2} .

Combining them yields

J = e - 14 + 19 e^{- 1/2} .

The upper

z

bound is

e^{y}

, as in the original PDF; an additional truncated version in the OCR must not be used.

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