Question

Integral $\int_{0}^{2}\int_{y^{2}}^{y+2} dxdy$ equals

• A. $\int_{0}^{1}\int_{0}^{\sqrt{x}}dydx+\int_{1}^{2}\int_{x=2}^{\sqrt{x}}dy~dx$
• B. $\int_{0}^{2}\int_{0}^{\sqrt{x}}dydx+\int_{2}^{4}\int_{x=2}^{x}dydx$
• C. $\int_{0}^{2}\int_{0}^{\sqrt{x}}dydx+\int_{2}^{4}\int_{x-2}^{\sqrt{x}}dy~dx$
• D. $\int_{0}^{2}\int_{0}^{\sqrt{x}}dydx+\int_{2}^{4}\int_{0}^{4\sqrt{x}}dydx$

Hint:

To change the order of integration, always plot the boundaries. If any "lower" or "upper" boundary curve changes its formula halfway through the range, you must split the integral into multiple parts.

Answer 1

The Correct Option is C

Changing the Order of Integration

Step 1: Original limits
The original integral is in the order \(dx\,dy\):
- Inner limits for \(x\): \(y^2 \le x \le y + 2\)
- Outer limits for \(y\): \(0 \le y \le 2\)
This describes a region bounded by the parabola \(x = y^2\) and the line \(x = y + 2\).

Step 2: Find corner points
- At \(y=0\): \(x = 0^2 = 0\), \(x = 0 + 2 = 2\) → points \((0,0)\) and \((2,0)\)
- At \(y=2\): \(x = 2^2 = 4\), \(x = 2 + 2 = 4\) → point \((4,2)\)
- Total range for \(x\): \(0 \le x \le 4\)

Step 3: Determine \(y\)-limits for each \(x\)
- For the parabola: \(x = y^2 \implies y = \sqrt{x}\)
- For the line: \(x = y + 2 \implies y = x - 2\)
Observing the region:
1. \(0 \le x \le 2\): \(y\) goes from 0 (x-axis) to \(\sqrt{x}\) (parabola)
2. \(2 \le x \le 4\): \(y\) goes from \(x-2\) (line) to \(\sqrt{x}\) (parabola)

Step 4: Write the integral in the order \(dy\,dx\)
\[ \int_{0}^{2} \int_{0}^{\sqrt{x}} dy\, dx + \int_{2}^{4} \int_{x-2}^{\sqrt{x}} dy\, dx \]

This corresponds to Option (3).