Question3) Find the area of the region bounded by $f(x)=2 e^{2 x}, g(x)=e^{2 x}+1$ and $[-1,2]$. (a) Find points of intersection/s. (b) In the interval $[-1,2]$ find which integral is above and which is below. (c) Write the integral to find the area inside $[-1,2], f(x)=2 e^{2 x}$ and $g(x)=e^{2 x}+1$. No need to integrate.

Studdy Solution
Set up the integral to find the area between $f(x) = 2e^{2x}$ and $g(x) = e^{2x} + 1$ over the interval $[-1, 2]$.
The area $A$ is given by:
$$A = \int_{-1}^{0} (g(x) - f(x)) \, dx + \int_{0}^{2} (f(x) - g(x)) \, dx$$
Substitute $f(x)$ and $g(x)$:
$$A = \int_{-1}^{0} \left( (e^{2x} + 1) - 2e^{2x} \right) \, dx + \int_{0}^{2} \left( 2e^{2x} - (e^{2x} + 1) \right) \, dx$$ $$A = \int_{-1}^{0} \left( 1 - e^{2x} \right) \, dx + \int_{0}^{2} \left( e^{2x} - 1 \right) \, dx$$
No need to integrate further as per the instructions.