UPSC 2026 Maths Optional Paper 1 Q3b-i — Step-by-Step Solution

10 marks · Section A

Double integrals · Calculus · asked 11× in 14 yrs · Read the full method →

Question

Evaluate the double integral $\displaystyle\int_{0}^{2}\int_{x^2+1}^{2x+1} x^2 y \,dy\,dx$ by reversing the order of integration.

Technique

Change of order of integration: sketch/describe the region $R$ from the given vertical-strip limits, find the corner points by intersecting $y=x^2+1$ with $y=2x+1$ , then re-cut $R$ by horizontal strips — solving each boundary curve for $x$ as a function of $y$ — and integrate $dx$ first. The reversal itself is the marked content, so the region description must be explicit.

Solution

Step 1 — Read off the region from the given limits.

The iterated integral, as printed, uses vertical strips:

R=\Big\{(x,y)\ :\ 0\le x\le 2,\ \ x^2+1\ \le\ y\ \le\ 2x+1\Big\}.

So $R$ is bounded below by the parabola $y=x^2+1$ and above by the line $y=2x+1$ .

Step 2 — Locate the corners (intersection of the two boundary curves).

x^2+1=2x+1 \;\Longrightarrow\; x^2-2x=0 \;\Longrightarrow\; x(x-2)=0 \;\Longrightarrow\; x=0,\ 2 .

The curves meet at $(0,1)$ and $(2,5)$ . Consistently, $2x+1-(x^2+1)=x(2-x)\ge 0$ exactly on $0\le x\le 2$ , so the line does lie above the parabola throughout the strip range and $R$ is the single closed lens-shaped region between them, with $1\le y\le 5$ .

Step 3 — Re-cut $R$ by horizontal strips (the reversal).

Fix $y$ with $1\le y\le 5$ and solve each boundary for $x$ :

Parabola: $y=x^2+1 \Rightarrow x^2=y-1 \Rightarrow x=\sqrt{y-1}$ (take the $+$ branch, since $x\ge 0$ on $R$ ).
Line: $y=2x+1 \Rightarrow x=\dfrac{y-1}{2}$ .

Now decide which is the left boundary. A point of $R$ satisfies both defining inequalities:

x^2+1\le y \iff x \le \sqrt{y-1}, \qquad y \le 2x+1 \iff x \ge \frac{y-1}{2}.

Hence for fixed $y$ ,

\frac{y-1}{2}\ \le\ x\ \le\ \sqrt{y-1},

i.e. the line is the left boundary and the parabola the right boundary. (This strip is non-empty: putting $u=\sqrt{y-1}\in[0,2]$ , the condition $\frac{u^2}{2}\le u$ is $u\le 2$ , true for all $y\in[1,5]$ , with equality only at $y=1,5$ — the two corners.)

Therefore

I=\int_{0}^{2}\!\!\int_{x^2+1}^{2x+1}\! x^2y\,dy\,dx \;=\; \int_{y=1}^{5}\int_{x=\frac{y-1}{2}}^{\sqrt{y-1}} x^2 y\;dx\,dy .

Step 4 — Inner integral (now in $x$ ).

\int_{\frac{y-1}{2}}^{\sqrt{y-1}} x^2\,dx=\left[\frac{x^3}{3}\right]_{\frac{y-1}{2}}^{\sqrt{y-1}}=\frac13\left[(y-1)^{3/2}-\frac{(y-1)^3}{8}\right].

I=\frac13\int_{1}^{5} y\left[(y-1)^{3/2}-\frac{(y-1)^3}{8}\right]dy .

Step 5 — Outer integral: substitute $t=y-1$ .

With $t=y-1$ , $y=t+1$ , $dy=dt$ , and $t:0\to 4$ ,

I=\frac13\int_{0}^{4}(t+1)\left(t^{3/2}-\frac{t^{3}}{8}\right)dt=\frac13\int_{0}^{4}\left(t^{5/2}+t^{3/2}-\frac{t^{4}}{8}-\frac{t^{3}}{8}\right)dt .

Antidifferentiate:

I=\frac13\left[\frac{2}{7}t^{7/2}+\frac{2}{5}t^{5/2}-\frac{t^{5}}{40}-\frac{t^{4}}{32}\right]_{0}^{4}.

At $t=4$ : $\;4^{7/2}=(2^2)^{7/2}=2^{7}=128$ , $\;4^{5/2}=2^{5}=32$ , $\;4^{5}=1024$ , $\;4^{4}=256$ . Hence

\frac{2}{7}(128)+\frac{2}{5}(32)-\frac{1024}{40}-\frac{256}{32}=\frac{256}{7}+\frac{64}{5}-\frac{128}{5}-8=\frac{256}{7}-\frac{64}{5}-8 .

Over the common denominator $35$ :

=\frac{1280-448-280}{35}=\frac{552}{35}.

Therefore

I=\frac13\cdot\frac{552}{35}=\frac{552}{105}=\frac{184}{35}.

Step 6 — Cross-check in the original order.

I=\int_0^2 x^2\left[\frac{y^2}{2}\right]_{x^2+1}^{2x+1}dx=\frac12\int_0^2 x^2\Big[(2x+1)^2-(x^2+1)^2\Big]dx.

Now $(2x+1)^2-(x^2+1)^2=(4x^2+4x+1)-(x^4+2x^2+1)=-x^4+2x^2+4x$ , so

I=\frac12\int_0^2\big(-x^{6}+2x^{4}+4x^{3}\big)dx=\frac12\left[-\frac{x^{7}}{7}+\frac{2x^{5}}{5}+x^{4}\right]_0^2=\frac12\left[-\frac{128}{7}+\frac{64}{5}+16\right].

=\frac12\cdot\frac{-640+448+560}{35}=\frac12\cdot\frac{368}{35}=\frac{184}{35}.\ \checkmark

Both orders agree, which is exactly the confirmation that the reversal of limits was done correctly.

Answer

\boxed{\;\int_{0}^{2}\!\int_{x^2+1}^{2x+1}\! x^{2}y\,dy\,dx=\int_{1}^{5}\!\int_{\frac{y-1}{2}}^{\sqrt{y-1}}\! x^{2}y\,dx\,dy=\frac{184}{35}\approx 5.2571\;}