UPSC 2022 Maths Optional Paper 1 Q1d — Step-by-Step Solution

10 marks · Section A

Improper integrals (unbounded interval/integrand) · Calculus · asked 5× in 14 yrs · Read the full method →

Question

Examine the convergence of $\displaystyle\int_0^2\dfrac{dx}{2x-x^2}$ .

Improper integral with two singular endpoints; limit comparison test against $\int dx/x$ ; direct integration via partial fractions to confirm.

Setup. $2x-x^2=x(2-x)$ . So the integrand $1/[x(2-x)]$ has singularities at both endpoints $x=0$ and $x=2$ .

The integral is improper at both endpoints. Split:

I=\int_0^1\dfrac{dx}{x(2-x)}+\int_1^2\dfrac{dx}{x(2-x)}.

Near $x=0$ : $\dfrac{1}{x(2-x)}\sim\dfrac{1}{2x}$ .

Compare with $\int_0^1\dfrac{dx}{x}$ , which diverges (logarithmic divergence at 0).

By limit comparison: $\lim_{x\to 0^+}\dfrac{1/[x(2-x)]}{1/x}=\lim_{x\to 0^+}\dfrac{1}{2-x}=\dfrac{1}{2}>0$ .

So $\int_0^1\dfrac{dx}{x(2-x)}$ diverges by limit comparison.

Since one piece diverges, the whole integral diverges.

\boxed{\;\int_0^2\dfrac{dx}{2x-x^2}\text{ diverges.}\;}