Animated Solution for Mathematics - Definite Integration: 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 :

Imagine standing on a coordinate plane, looking at two distinct paths. One is the elegant, sweeping curve of the parabola $y=x^2$, and the other is the sharp, descending slope of the hyperbola $y=\frac{1}{x}$.

These two curves are not just lines on a graph; they are the boundaries of a hidden territory. Our mission is to find the value of $t$ such that the area of the region bounded by these curves, the x-axis ($y=0$), and the vertical line $x=t$ is exactly $1$ square unit.

Before we can measure the area, we must find where these two paths meet. This is our pivot point. By setting $x^2=\frac{1}{x}$, we are essentially asking where these two worlds collide.

Multiplying both sides by $x$ gives us $x^3=1$, which leads us directly to $x=1$. At this point, both curves share the same height, $y=1$.

This point $x=1$ is crucial because it is where the "upper boundary" of our region switches from the parabola to the hyperbola.

Now, let's visualize the region. From $x=0$ to $x=1$, the parabola $y=x^2$ sits above the x-axis, forming the ceiling of our region.

As soon as we cross $x=1$, the hyperbola $y=\frac{1}{x}$ takes over as the ceiling. Because the "ceiling" changes, we cannot use a single integral.

We must split our journey into two distinct phases. Phase 1 is the area under the parabola from $0$ to $1$, and Phase 2 is the area under the hyperbola from $1$ to $t$.

Let's calculate the area of Phase 1, which we will call $A_1$. We set up the integral:

Integrating $x^2$ gives us $\frac{x^3}{3}$. Evaluating this from $0$ to $1$, we get:

Now for Phase 2, $A_2$. This is the area under the hyperbola from $1$ to $t$:

The integral of $\frac{1}{x}$ is the natural logarithm, $\ln (x)$. Evaluating this from $1$ to $t$, we get $\ln (t)-\ln (1)$. Since $\ln (1)=0$, our second area is simply $\ln (t)$.

The total area is the sum of these two parts:

The problem states this total area is $1$. So, we set up our final equation:

Subtracting $\frac{1}{3}$ from both sides, we find:

To isolate $t$, we use the inverse of the natural logarithm, which is the exponential function. Raising $e$ to the power of both sides, we get: