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

This question asks us to redraw the region of integration for $I$ and read off the new order $dy\,dx$. Let's identify the exact region traced by the two given pieces and check each option by matching both the region description and the numeric value of $I$.

The first piece $\int_0^1\int_0^{\sqrt y} dx\,dy$ covers all points with $0\le x\le\sqrt y$ and $0\le y\le1$; squaring the bound $x\le\sqrt y$ gives $x^2\le y$, so this piece is the region between the parabola $y=x^2$ and the line $y=1$, for $0\le x\le1$.

The second piece $\int_1^2\int_{\sqrt{y-1}}^1 dx\,dy$ covers points with $\sqrt{y-1}\le x\le1$ and $1\le y\le2$; squaring gives $y-1\le x^2$, i.e. $y\le x^2+1$, so this piece is the region between $y=1$ and $y=x^2+1$, again for $0\le x\le1$.

The two pieces meet exactly at $y=1$, so together they form one continuous region: for every $x$ from $0$ to $1$, $y$ runs from $x^2$ up to $x^2+1$. No split at $x=\tfrac12$ is needed.

1. Option A: uses bounds $-x$ to $x$ and $-x^2$ to $-x^2+1$, a completely different pair of curves that does not match either piece above. Direct evaluation gives $\tfrac34$, not the true value of $I$.
2. Option B: $\int_0^1\int_{x^2}^{x^2+1}dy\,dx$, exactly the single merged region found above, with no artificial split needed.
3. Option C: splits the region at $x=\tfrac12$ and uses $y=\sqrt x$ and $y=\sqrt{x+1}$ in the second piece, curves that never appeared in the original description. It evaluates to an irrational number, not matching $I$.
4. Option D: $\int_0^1\int_{x^2}^{x^2-1}dy\,dx$ has an upper limit $x^2-1$ smaller than the lower limit $x^2$, an impossible negative height for a real region, and it evaluates to $-1$, the wrong sign entirely.

Computing the original integral directly confirms the check: $\int_0^1\sqrt y\,dy=\tfrac23$ and $\int_1^2\left(1-\sqrt{y-1}\right)dy=\tfrac13$, so $I=\tfrac23+\tfrac13=1$. Evaluating option B the same way, $\int_0^1\left[(x^2+1)-x^2\right]dx=\int_0^1 1\,dx=1$, the same value.

Let's summarize:

• The two given pieces are the regions between $y=x^2$ and $y=1$, and between $y=1$ and $y=x^2+1$, which join into one region $x^2\le y\le x^2+1$ for $0\le x\le1$.
• Only option B reproduces this region and the correct value $I=1$; the other three either use the wrong bounding curves or an impossible negative height.

So the correct expression after changing the order of integration is $\int_0^1\int_{x^2}^{x^2+1}dy\,dx$, option B.

\[ \boxed{I = \int_0^1 \int_{x^2}^{x^2+1} dy\, dx} \]