The area (in sq. units) of the region bounded by the curves $$y = 2^x$$ and $$y = x + 1$$, in the first quadrant is

We are asked to find the area, in square units, enclosed by the two curves

$$y = 2^{x} \quad\text{and}\quad y = x + 1$$

in the first quadrant. “First quadrant’’ means we restrict ourselves to points where $$x \ge 0$$ and $$y \ge 0$$.

To set up the required definite integral, we first need the points where the two graphs meet. Intersection points are obtained by equating the two expressions for $$y$$:

$$2^{x} = x + 1.$$

We solve this equation step by step.

• Try $$x = 0$$: $$2^{0} = 1,\; x+1 = 0+1 = 1.$$   Both sides are equal, so $$x = 0$$ is a solution.

• Try $$x = 1$$: $$2^{1} = 2,\; x+1 = 1+1 = 2.$$   Again both sides are equal, so $$x = 1$$ is also a solution.

To check whether there are any further solutions for $$x \ge 0$$, consider the function

$$f(x) = 2^{x} - (x + 1).$$

We have already found $$f(0) = 0$$ and $$f(1) = 0$$. Differentiate to see the behaviour between and beyond these points:

$$f'(x) = (\ln 2)\,2^{x} - 1.$$

At $$x = 0$$, $$f'(0) = (\ln 2)\cdot 1 - 1 = \ln 2 - 1 \lt 0.$$ At $$x = 1$$, $$f'(1) = (\ln 2)\cdot 2 - 1 = 2\ln 2 - 1 \gt 0.$$

Since $$f'(x)$$ changes from negative to positive only once, $$f(x)$$ can cross the $$x$$-axis at most twice. We have already identified two crossings, so those are the only intersections in the first quadrant:

$$\bigl(0,\,1\bigr)\quad\text{and}\quad\bigl(1,\,2\bigr).$$

Next, we decide which curve lies above the other between these two $$x$$-values. Pick a convenient test point such as $$x = 0.5$$:

$$y = 2^{0.5} = \sqrt{2} \approx 1.414,$$ $$y = 0.5 + 1 = 1.5.$$

Clearly $$1.5 \gt 1.414,$$ so throughout the interval $$0 \le x \le 1$$ the straight line $$y = x + 1$$ is the upper curve and the exponential curve $$y = 2^{x}$$ is the lower curve.

The standard formula for the area between two curves, when the region is projected on the $$x$$-axis from $$x = a$$ to $$x = b$$, is

$$A = \int_{a}^{b} \bigl[y_{\text{upper}} - y_{\text{lower}}\bigr]\,dx.$$

In our case:

$$a = 0, \quad b = 1,$$ $$y_{\text{upper}} = x + 1, \quad y_{\text{lower}} = 2^{x}.$$

Hence

$$ \begin{aligned} A &= \int_{0}^{1} \Bigl[(x + 1) - 2^{x}\Bigr]\,dx \\[4pt] &= \int_{0}^{1} (x + 1)\,dx \;-\; \int_{0}^{1} 2^{x}\,dx. \end{aligned} $$

We evaluate each integral separately.

For the first integral, recall the power-rule of integration:

$$\int x^{n}\,dx = \frac{x^{n+1}}{n+1} + C.$$

Applying it term by term,

$$ \int_{0}^{1} (x + 1)\,dx = \int_{0}^{1} x\,dx + \int_{0}^{1} 1\,dx = \left[\frac{x^{2}}{2}\right]_{0}^{1} + \left[x\right]_{0}^{1} = \left(\frac{1^{2}}{2} - \frac{0^{2}}{2}\right) + (1 - 0) = \frac{1}{2} + 1 = \frac{3}{2}. $$

For the second integral, we recall the exponential-base-$$2$$ integration rule:

$$\int 2^{x}\,dx = \frac{2^{x}}{\ln 2} + C,$$

because differentiating $$\dfrac{2^{x}}{\ln 2}$$ gives back $$2^{x}$$.

Therefore,

$$ \int_{0}^{1} 2^{x}\,dx = \left[\frac{2^{x}}{\ln 2}\right]_{0}^{1} = \frac{2^{1}}{\ln 2} - \frac{2^{0}}{\ln 2} = \frac{2}{\ln 2} - \frac{1}{\ln 2} = \frac{1}{\ln 2}. $$

Substituting the two evaluated parts back into the expression for $$A$$, we obtain

$$ A = \frac{3}{2} - \frac{1}{\ln 2}. $$

This numerical expression exactly matches Option A.

Hence, the correct answer is Option A.