Question

Let \( I \) be the integral defined as follows:
\[ I = \int_0^1 \int_0^{\sqrt{y}} dx\, dy \;+\; \int_1^2 \int_{\sqrt{y-1}}^{1} dx\, dy \]
If the order of integration is changed, then which one of the following is the correct expression for \( I \)?

• A. \[ \int_0^{1/2} \int_{-x}^{x} dy\, dx \;+\; \int_{1/2}^{1} \int_{-x^2}^{-x^2+1} dy\, dx \]
• B. \[ \int_0^1 \int_{x^2}^{x^2+1} dy\, dx \]
• C. \[ \int_0^{1/2} \int_{x^2}^{x^2+1} dy\, dx \;+\; \int_{1/2}^{1} \int_{\sqrt{x}}^{\sqrt{x+1}} dy\, dx \]
• D. \[ \int_0^1 \int_{x^2}^{x^2-1} dy\, dx \]

Hint:

Sketch both original regions in the xy-plane and see exactly where they join along the line y = 1.

Answer 1

The Correct Option is B

Step 1: Understanding the Question.
We are given \( I \) as the sum of two double integrals in the order \( dx\,dy \), and we need to sketch the region each one covers, then rewrite the same region in the order \( dy\,dx \).

Step 2: Key Formula or Approach.
For a region described by \( 0\le x\le\sqrt{y} \), squaring the inequality gives \( x^2\le y \), which describes the same region using a curve in \( x \). We use this trick on both pieces, then check where the two pieces join so we can write the combined region as a single integral in \( x \) first.

Step 3: Detailed Explanation.
First piece: \( 0\le y\le1 \), \( 0\le x\le\sqrt{y} \). Squaring \( x\le\sqrt{y} \) gives \( x^2\le y \). Since \( y\le1 \) and \( x\le\sqrt{y}\le1 \), this region is exactly \( x^2\le y\le1 \) for \( 0\le x\le1 \).
Second piece: \( 1\le y\le2 \), \( \sqrt{y-1}\le x\le1 \). Squaring \( \sqrt{y-1}\le x \) gives \( y-1\le x^2 \), i.e. \( y\le x^2+1 \). Combined with \( y\ge1 \), this region is exactly \( 1\le y\le x^2+1 \) for \( 0\le x\le1 \) (since \( x^2+1\le2 \) whenever \( x\le1 \)).
The two pieces share the boundary \( y=1 \), so together they form one continuous region: for every \( x \) from \( 0 \) to \( 1 \), \( y \) runs from \( x^2 \) up to \( x^2+1 \), with no need to split the range of \( x \). So
\[ I = \int_0^1 \int_{x^2}^{x^2+1} dy\, dx \]
which is option (B). As a check, compute the value both ways. Directly, \( \int_0^1 \sqrt{y}\,dy = \tfrac{2}{3} \) and \( \int_1^2 \left(1-\sqrt{y-1}\right)dy = \tfrac{1}{3} \), giving \( I = \tfrac{2}{3}+\tfrac{1}{3}=1 \). From option (B), \( \int_0^1\left[(x^2+1)-x^2\right]dx = \int_0^1 1\,dx = 1 \), the same value.
Option (A) uses the curves \( y=-x \), \( y=x \) and \( y=-x^2 \), \( y=-x^2+1 \), which never appear in the original region description; it evaluates to \( \tfrac{3}{4} \), not \( 1 \). Option (C) splits at \( x=\tfrac12 \) and brings in the curves \( y=\sqrt{x} \), \( y=\sqrt{x+1} \), also not part of the original region; it evaluates to an irrational number, not \( 1 \). Option (D) has upper limit \( x^2-1 \) below the lower limit \( x^2 \), an impossible negative height for a genuine region, and it evaluates to \( -1 \), the wrong sign.

Step 4: Final Answer.
The single merged region \( x^2\le y\le x^2+1 \), \( 0\le x\le1 \), reproduces both the shape and the value of \( I \).
\[ \boxed{I = \int_0^1 \int_{x^2}^{x^2+1} dy\, dx = 1} \]