Question

If \( \displaystyle \int_0^1 \int_{2y}^2 e^{x^2} \, dx \, dy = k(e^4 - 1), \) then \( k \) equals ...............

Hint:

Always visualize integration limits before changing the order; it simplifies double integrals.

Answer 1

Correct Answer: 0.25

Step 1: Express the given integral.
\[ I = \int_0^1 \int_{2y}^2 e^{x^2} \, dx \, dy. \]

Step 2: Change the order of integration.
The region is bounded by \( 0 \le y \le 1 \) and \( 2y \le x \le 2. \) Inverting limits: \( 0 \le x \le 2, \; 0 \le y \le \frac{x}{2}. \) \[ I = \int_0^2 \int_0^{x/2} e^{x^2} \, dy \, dx = \int_0^2 e^{x^2} \left( \frac{x}{2} \right) dx = \frac{1}{2} \int_0^2 x e^{x^2} dx. \]

Step 3: Integrate.
Let \( t = x^2 $\Rightarrow$ dt = 2x dx $\Rightarrow$ x dx = \frac{dt}{2}. \) \[ I = \frac{1}{2} \cdot \frac{1}{2} \int_0^4 e^t dt = \frac{1}{4}(e^4 - 1). \] Hence \( k = \frac{1}{4}. \)

Final Answer: \[ \boxed{k = \frac{1}{4}} \]