Discussion
Let I be the integral defined as follows:
\[ \begin{aligned} I = &\int_0^1 \int_0^{\sqrt{y}} dx\, dy \\ &+ \int_1^2 \int_{\sqrt{y-1}}^1 dx\, dy \end{aligned} \]
If the order of the integration is changed, then which one of the following is the correct expression for I?
Answer: B
The answer is \(\int_0^1 \int_{x^2}^{x^2+1} dy\, dx\).
Read off the region. In the first integral \(0 \le y \le 1\) and \(0 \le x \le \sqrt{y}\), that is \(y \ge x^2\). In the second, \(1 \le y \le 2\) and \(\sqrt{y - 1} \le x \le 1\), that is \(y \le x^2 + 1\). Together they cover, for each x in [0,1], the strip between the two parabolas:
\[ 0 \le x \le 1, \quad x^2 \le y \le x^2 + 1 \]
so a single integral in the order dy dx suffices. The option with upper limit \(x^2 - 1\) has the limits the wrong way round (it gives a negative area), and the two split options use boundaries such as \(y = \pm x\) and \(y = \sqrt{x}\) that are not edges of this region.
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