Question

The value of the integral \[ \int_0^1 \int_{y^2}^1 \frac{e^x}{\sqrt{x}} \, dx \, dy \] equals ............ (round off to two decimal places):

Hint:

Always visualize the region before interchanging limits in double integrals — it simplifies the evaluation drastically.

Answer 1

Correct Answer: 1.7

Step 1: Change the order of integration.
Region: $0 \le y \le 1$, $y^2 \le x \le 1$. Equivalent: $0 \le x \le 1$, $0 \le y \le \sqrt{x}$.

Step 2: Integrate with respect to $y$ first.
\[ I = \int_0^1 \frac{e^x}{\sqrt{x}} \left[\int_0^{\sqrt{x}} dy \right] dx = \int_0^1 \frac{e^x}{\sqrt{x}} \cdot \sqrt{x} \, dx = \int_0^1 e^x \, dx. \]

Step 3: Integrate over $x$.
\[ I = e^x \Big|_0^1 = e - 1 \approx 2.718 - 1 = 1.718. \] Wait, recheck range: actual inner integral yields factor $(\sqrt{x} - 0)$ — correct. So, \[ I = e - 1 = 1.72. \]

Step 4: Rounding to two decimal places.
\[ \boxed{1.72.} \]