 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-2/19/a/solution

The Leibniz integral rule accounts for both the moving upper endpoint and the integrand's dependence on

a

I^{'} (a) = f (a, a) + \int_{0}^{a} \frac{\partial f}{\partial a} (x, a) d x .

For the specified integrand this gives

I^{'} (a) = e^{- a^{2}} - \int_{0}^{a} x e^{- a x} d x

. When

a \neq = 0

\int_{0}^{a} x e^{- a x} d x = - e^{- a^{2}} + \frac{1 - e ^{- a^{2}}}{a ^{2}},

I^{'} (a) = 2 e^{- a^{2}} - \frac{1 - e ^{- a^{2}}}{a ^{2}} (a \neq = 0) .

For the requested independent check, direct integration gives

I (a) = (1 - e^{- a^{2}}) / a

a \neq = 0

[2 a^{2} e^{- a^{2}} - (1 - e^{- a^{2}})] / a^{2}

, agreeing with the result above. At zero the apparent singularity is removable:

I (a) = a - a^{3} /2 + O (a^{5})

, so

I (0) = 0, I^{'} (0) = 1

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