Answer
$\dfrac{5}{2}-\dfrac{1}{\ln 2}$
Work Step by Step
Let us consider that two continuous functions $f(x)$ and $g(x)$ with $f(x)\geq g(x)$ on the interval $[a,b]$ . The area (A) of the region bounded by the graph of $f(x)$ and $g(x)$ on the interval $[a,b]$ can be calculated as: $Area(A)=\int_a^b [f(x)-g(x)] \ dx$
The given curves are $y=2^x$ and $y=3-x$.The intersection points of these curves satisfy the given equations are: $2^x=3-x \implies x=0 \ \text{and}\ x=1$ . These points will be the lower and upper limit of the integration.
Thus, the area of the region is:
$A=\int_a^b [f(x)-g(x)] \ dx= \int_{0}^{1} [3-x-2^x] \ dx\\ =[3x-\dfrac{x^2}{2}-\dfrac{2^x}{\ln 2}]_{0}^1\\ = (3-\dfrac{1^2}{2}-\dfrac{2^1}{\ln 2})-(3(0)-\dfrac{0^2}{2}-\dfrac{2^0}{\ln 2})\\=\dfrac{5}{2}-\dfrac{1}{\ln 2}$