Key Concepts and Formulas

• The area under a curve $y = f(x)$ from $x = a$ to $x = b$, where $f(x) \ge 0$ on $[a, b]$, is given by the definite integral: $A = \int_a^b f(x) \, dx$
• The area between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$, where $f(x) \ge g(x)$ on $[a, b]$, is given by the definite integral: $A = \int_a^b (f(x) - g(x)) \, dx$
• The integral of $x^n$ is: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$
• The integral of $1/x$ is: $\int \frac{1}{x} \, dx = \ln|x| + C$

Step-by-Step Solution

Step 1: Visualize the Region and Find Intersection Points

We need to find the area of the region bounded by $y = x^2$, $y = \frac{1}{x}$, $y = 0$, and $x = t$ (where $t > 1$). First, let's find the intersection point of $y = x^2$ and $y = \frac{1}{x}$. We set the expressions for $y$ equal to each other to find where they intersect because that point will be crucial for setting up our integrals.

$x^2 = \frac{1}{x}$ $x^3 = 1$ $x = 1$

When $x = 1$, $y = 1^2 = 1$. So, the intersection point is $(1, 1)$. This is important because the region's boundaries change at this point.

Step 2: Decompose the Area into Integrable Parts

The region is bounded by the x-axis ($y=0$) from below. Since $t > 1$, we need to split the area into two parts based on the intersection point at $x = 1$.

• Part 1 ($A_1$): From $x = 0$ to $x = 1$. The upper boundary is $y = x^2$, and the lower boundary is $y = 0$. Thus, the area $A_1$ is given by: $A_1 = \int_0^1 x^2 \, dx$

• Part 2 ($A_2$): From $x = 1$ to $x = t$. The upper boundary is $y = \frac{1}{x}$, and the lower boundary is $y = 0$. Thus, the area $A_2$ is given by: $A_2 = \int_1^t \frac{1}{x} \, dx$

The total area $A$ is the sum of these two parts: $A = A_1 + A_2 = \int_0^1 x^2 \, dx + \int_1^t \frac{1}{x} \, dx$

Step 3: Evaluate the Definite Integrals

Now, we evaluate each integral.

• Evaluation of $A_1$: $A_1 = \int_0^1 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} - 0 = \frac{1}{3}$

• Evaluation of $A_2$: $A_2 = \int_1^t \frac{1}{x} \, dx = \left[\ln|x|\right]_1^t = \ln|t| - \ln|1| = \ln t - 0 = \ln t$ (Since $t>1$, we can write $\ln t$ without the absolute value).

Step 4: Calculate the Total Area and Solve for $t$

We are given that the total area $A = 1$. Therefore: $A = A_1 + A_2 = \frac{1}{3} + \ln t = 1$

Now, solve for $t$: $\ln t = 1 - \frac{1}{3} = \frac{2}{3}$ $t = e^{\frac{2}{3}}$

Common Mistakes & Tips

• Sketching is Crucial: Always sketch the region to identify the correct upper and lower boundaries and intersection points.
• Splitting the Integral: Remember to split the integral into multiple parts if the upper or lower boundary changes within the interval.
• Logarithm Properties: Recall that $\ln(1) = 0$.

Summary

By visualizing the region, splitting the area into two integrals, and evaluating each integral, we found that the value of $t$ that gives an area of 1 square unit is $e^{2/3}$.

The final answer is $\boxed{e^{2/3}}$, which corresponds to option (D).