Question 1: (1 point)
Find the area of the region bounded by the curves $f(x)=3 \cdot \sin \left(\frac{\pi}{4} x\right)$ and $g(x)=2^{x}-1$, and $x \geq 0$

Step 1 ：First, we need to find the intersection points of the two curves $f(x)=3 \cdot \sin \left(\frac{\pi}{4} x\right)$ and $g(x)=2^{x}-1$.

Step 2 ：To find the intersection points, we set $f(x) = g(x)$, which gives us $3 \cdot \sin \left(\frac{\pi}{4} x\right) = 2^{x}-1$.

Step 3 ：Solving this equation is not straightforward, so we will use a numerical method to find the approximate solutions. Using a graphing calculator or software, we find that the curves intersect approximately at $x = 0$ and $x = 1.2304$.

Step 4 ：The area between the curves is given by the integral of the absolute difference of the two functions from $x = 0$ to $x = 1.2304$.

Step 5 ：So, the area $A$ is given by $A = \int_{0}^{1.2304} |3 \cdot \sin \left(\frac{\pi}{4} x\right) - (2^{x}-1)| dx$.

Step 6 ：We can split this integral into two parts, from $0$ to $1$ where $f(x) > g(x)$, and from $1$ to $1.2304$ where $g(x) > f(x)$.

Step 7 ：So, $A = \int_{0}^{1} (3 \cdot \sin \left(\frac{\pi}{4} x\right) - (2^{x}-1)) dx + \int_{1}^{1.2304} ((2^{x}-1) - 3 \cdot \sin \left(\frac{\pi}{4} x\right)) dx$.

Step 8 ：Calculating these integrals using a numerical method, we find that $A \approx 0.893$.

Step 9 ：\boxed{A \approx 0.893}$ is the area of the region bounded by the curves $f(x)=3 \cdot \sin \left(\frac{\pi}{4} x\right)$ and $g(x)=2^{x}-1$, and $x \geq 0$.