91Ó°ÊÓ

Finding the area under a curve is a crucial concept in calculus, especially when dealing with parametric equations. A parametric equation defines a curve using a parameter, typically represented by \(t\), instead of an explicit function \(y = f(x)\). In our exercise, the parametric equations are \(x = t^2\) and \(y = \ln(t)\). These equations express the coordinates \((x, y)\) in terms of a third variable \(t\), which is confined within a specific interval. In this case, \(0 \leq t \leq e\).

When the task is to find the area under such a curve, the strategy involves setting up an integral that covers the given interval of \(t\). The standard formula used is:

• Area = \( \int_a^b y(t) \frac{dx}{dt} \, dt \)

Here, \(y(t)\) is the function in terms of \(t\), and \(\frac{dx}{dt}\) is the derivative of \(x(t)\). This integral effectively adds up infinitely small slices of area beneath the curve from \(t = a\) to \(t = b\).

This approach converts the complex geometric shape into a more manageable mathematical form, which can then be computed to find the total area under the curve.

Integration by parts is a technique used to integrate products of functions. It's particularly handy when simple integration techniques don't suffice. The formula is derived from the product rule for differentiation and is written as:

• \( \int u \, dv = uv - \int v \, du \)

Here, \(u\) and \(dv\) are parts of the integrand where \(u = f(t)\) is a function we'll differentiate, and \(dv = g'(t) \, dt\) is a function we'll integrate. Choosing \(u\) and \(dv\) can be strategic to simplify the process.

In our scenario, evaluating \( \int t \ln(t) \, dt \) involves selecting \(u = \ln(t)\) and \(dv = t \, dt\). Differentiate \(u\) to find \(du = \frac{1}{t} \, dt\) and integrate \(dv\) to get \(v = \frac{t^2}{2}\). Plug these back into the integration by parts formula to break down the integral into manageable pieces.

This method often transforms a complex integral into a far simpler one, making it possible to solve integrals that appear intractable at first glance.

A definite integral calculates the total accumulation of quantities along a curve from one point to another. For a definite integral

• \( \int_a^b f(t) \, dt \)

\(a\) and \(b\) represent the limits of integration, which specify the beginning and end of the interval over which we're integrating. In our case, these limits are \(0\) and \(e\), respectively.

To evaluate a definite integral, compute the indefinite integral first, and then apply the limits to find the net area. The Fundamental Theorem of Calculus underpins this process. Once the indefinite integral, representing a function's antiderivative, is obtained, evaluate it from the upper limit to the lower limit:

• \([F(b) - F(a)]\)

In the problem, solving the integral gives the expression \([ \frac{t^2 \ln(t)}{2} - \frac{t^2}{4} ]_0^e\). Computing this evaluates the area under the curve explicitly from \(t=0\) to \(t=e\), leading to the final answer. This process converts abstract mathematical expressions into concrete numerical values, giving a clear geometric interpretation of the integral as an area.