Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 20 / a

Table of contents
- Solution a

Solution

For

α \neq = 0

, integration by substitution

t = αx

gives

I (α) = \int_{0}^{1} \frac{d x}{1 + α ^{2} x ^{2}} = \frac{arctan α}{α} .

At

α = 0

the integrand is one, so

I (0) = 1

, also the continuous limit of this expression. For nonzero

α

, differentiation under the integral sign is valid on the finite interval and yields

I^{'} (α) = - 2 α \int_{0}^{1} \frac{x ^{2}}{( 1 + α ^{2} x ^{2} ) ^{2}} d x = \frac{α / ( 1 + α ^{2} ) - a r c t a n α}{α ^{2}} .

α = 3

and use

arctan 3 = π /3

. The requested value is

\int_{0}^{1} \frac{x ^{2}}{( 1 + 3 x ^{2} ) ^{2}} d x = - \frac{I ^{'} ( 3 )}{2 3} = \frac{π}{18 3} - \frac{1}{24} = \frac{π 3}{54} - \frac{1}{24} .

 Ancestors (12)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version