← 2017 Paper 1

UPSC 2017 Maths Optional Paper 1 Q4c — 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 if the improper integral $\displaystyle\int_0^3 \frac{2x\,dx}{(1-x^2)^{2/3}}$ exists.

Technique

Identify the interior singularity at $x=1$, split the integral, compare the singularity order ($|1-x|^{-2/3}$, exponent $<1$) to confirm convergence, then evaluate with the real-cube-root antiderivative $-3(1-x^2)^{1/3}$.

Solution

Step 1 — Locate the singularity

The integrand $\dfrac{2x}{(1-x^2)^{2/3}}$ blows up where $1-x^2=0$, i.e. at $x=1$, which is interior to $[0,3]$. (Note $(1-x^2)^{2/3}=\big((1-x^2)^2\big)^{1/3}\ge0$ is well-defined and positive for $x\ne1$, including $x>1$ where $1-x^2<0$.) Split:

each an improper integral with the singularity at the endpoint $x=1$.

Step 2 — Antiderivative

Let $u=1-x^2$, $du=-2x\,dx$. Then

interpreting $(\cdot)^{1/3}$ as the real cube root (so it is continuous through $x=1$).

Step 3 — Convergence near $x=1$ (order of the singularity)

Near $x=1$, $1-x^2=(1-x)(1+x)\sim 2(1-x)$, so the integrand behaves like $\dfrac{2}{(2(1-x))^{2/3}}=C\,|1-x|^{-2/3}$. Since the exponent $\tfrac23<1$, the integral of $|1-x|^{-2/3}$ converges near $x=1$. Both improper pieces therefore converge.

Step 4 — Evaluate

Using $F(x)=-3(1-x^2)^{1/3}$ (real cube root, continuous on $[0,3]$):

Hence

Answer