If the area of the region bounded by the curves, $$y = x^2$$, $$y = \frac{1}{x}$$ and the lines $$y = 0$$ and $$x = t$$ (t > 1) is 1 sq. unit, then t is equal to:

First we locate the point where the two given curves meet, because that point will divide the region into two natural parts. We set $$y=x^{2}$$ equal to $$y=\dfrac{1}{x}$$ and obtain

$$x^{2}=\frac{1}{x}\; \Longrightarrow\; x^{3}=1\; \Longrightarrow\; x=1.$$

Putting $$x=1$$ in either curve gives $$y=1,$$ so the curves intersect at the single point $$(1,1).$$ Because we are told that $$t\gt 1,$$ the vertical line $$x=t$$ lies to the right of this intersection.

The boundary therefore runs from the origin $$(0,0)$$ up the parabola $$y=x^{2}$$ to $$(1,1),$$ then along the hyperbola $$y=\dfrac{1}{x}$$ from $$x=1$$ to $$x=t,$$ and finally straight down the line $$x=t$$ to the point $$(t,0)$$ on the $$x$$-axis, after which the $$x$$-axis brings us back to the origin. Hence, above the $$x$$-axis the region is exactly the set of points lying under

$$y=x^{2}\quad \text{for }0\le x\le 1,$$

and under

$$y=\frac{1}{x}\quad \text{for }1\le x\le t.$$

To find the required area we add the area under each curve separately. The general formula for the area under a non-negative curve $$y=f(x)$$ from $$x=a$$ to $$x=b$$ is

$$\text{Area}=\int_{a}^{b}f(x)\,dx.$$

Applying this once for the parabola and once for the hyperbola, we write

$$\text{Area}= \int_{0}^{1}x^{2}\,dx + \int_{1}^{t}\frac{1}{x}\,dx.$$

We evaluate the first integral. Using the power rule $$\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C,$$ we get

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

For the second integral we recall the standard result $$\int \dfrac{1}{x}\,dx=\ln|x|+C.$$ Hence

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

Adding the two pieces we obtain

$$\text{Area}=\frac{1}{3}+\ln t.$$

The problem states that this area equals $$1$$ square unit, so we set

$$\frac{1}{3}+\ln t = 1.$$

Isolating $$\ln t$$ we subtract $$\dfrac{1}{3}$$ from both sides:

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

Now we exponentiate to solve for $$t$$. Because $$e^{\ln t}=t,$$ we have

$$t = e^{\,\frac{2}{3}}.$$

This value matches Option A.

Hence, the correct answer is Option A.