Problem 24 - Applied Calculus

Question

Find the exact area. $$\text { Under } y=t e^{-t} \text { for } 0 \leq t \leq 2$$

Step-by-Step Solution

Answer

The exact area is \( 1 - \frac{3}{e^2} \).

Step 1: Understand the Integral Representation

The area under the curve described by the function \( y = t e^{-t} \) from \( t = 0 \) to \( t = 2 \) is given by the definite integral \( \int_{0}^{2} t e^{-t} \, dt \). We need to compute this integral to find the exact area.

Step 2: Choose Integration Technique

To solve the integral \( \int_{0}^{2} t e^{-t} \, dt \), we observe that the integrand is a product of a polynomial \( t \) and an exponential function \( e^{-t} \). This suggests that integration by parts is the suitable technique to use.

Step 3: Apply Integration by Parts

The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). We choose \( u = t \) so that \( du = dt \), and \( dv = e^{-t} \, dt \) so that \( v = -e^{-t} \). Applying the formula:\[\int t e^{-t} \, dt = -t e^{-t} - \int -e^{-t} \, dt\] \[= -t e^{-t} + \int e^{-t} \, dt.\]

Step 4: Solve the Remaining Integral

The remaining integral is \( \int e^{-t} \, dt \), which is \(-e^{-t} + C\). Thus, the integral becomes:\[-t e^{-t} + (-e^{-t}) = -t e^{-t} - e^{-t}.\]

Step 5: Evaluate the Definite Integral

Now, substitute the limits of integration:\[\int_{0}^{2} t e^{-t} \, dt = \left[ -(t e^{-t} + e^{-t}) \right]_{0}^{2} = \left[ -(2 e^{-2} + e^{-2}) \right] - \left[ -(0 e^{0} + e^{0}) \right].\]Simplify to get:\[-(3e^{-2}) + 1 = 1 - \frac{3}{e^2}.\]

Step 6: Final Calculation

The constant \( e \approx 2.718 \), so its square \( e^2 \approx 7.389 \). Therefore, \( \frac{3}{e^2} \approx \frac{3}{7.389} \approx 0.406 \).So, the exact area is approximately:\[1 - 0.406 = 0.594.\]

Step 7: Write the Exact Result

In an exact, non-approximated form, the area under the curve \( y = t e^{-t} \) from \( t = 0 \) to \( t = 2 \) is:\[1 - \frac{3}{e^2}.\] This is the exact answer for the problem, acknowledging that numerical approximations were provided for clarity.

Key Concepts

Integration by Parts

Integration by parts is a fundamental technique used to solve integrals, especially when dealing with the product of two different types of functions, such as a polynomial and an exponential function. It is derived from the product rule for differentiation. The integration by parts formula is:\[ \int u \, dv = uv - \int v \, du \]To apply this formula, choose two parts from the integral: one to differentiate (\( u \)) and one to integrate (\( dv \)).

• Let's take \( u = t \), which simplifies upon differentiation to \( du = dt \).
• For \( dv = e^{-t} \, dt \), integration yields \( v = -e^{-t} \).

Substituting into the formula, it helps in turning the original integral into one that's easier to solve. This technique is a powerful tool, especially in cases like this exercise where you work with products of functions.

Exponential Function

The exponential function, particularly in the form \( e^x \) or \( e^{-x} \), plays a crucial role in calculus and differential equations. It models processes that change at rates proportional to their current value, which is common in natural processes like population growth or radioactive decay. In this exercise, we encounter the exponential function \( e^{-t} \).Some important properties of the exponential function include:

• Derivative: The derivative of \( e^{x} \) is \( e^{x} \), and similarly, the derivative of \( e^{-x} \) is \( -e^{-x} \).
• Integration: The integral of \( e^{x} \) is \( e^{x} + C \), and the integral of \( e^{-x} \) becomes \( -e^{-x} + C \).
• \( e \) as a base: The base of the natural exponential function is approximately 2.718, and this constant is widely respected in both pure and applied mathematics.

Polynomial

Polynomial functions, like \( t \) in this exercise, are expression involving a finite sum of terms, each of which is made up of a constant coefficient and a variable raised to a non-negative integer power. They are some of the simplest functions you can integrate or differentiate.A polynomial of degree \( n \) in variable \( x \) looks like this:\[ P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \]Where:

• \( n \) is the degree of the polynomial, indicating the highest power of \( x \).
• \( a_0, a_1, \ldots, a_n \) are coefficients.

For integration purposes, when you integrate \( x^n \), the result is:\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]In the context of this exercise, the polynomial \( t \) interacts with the exponential function in the integral, so it requires the application of integration by parts to resolve.

Area Under the Curve

Finding the area under a curve is a common application of integration in calculus. This represents the "accumulated value" between two points on the curve and along the x-axis, encapsulating the concept of net movement, accumulation of mass, or even cost with varying rates.For a given function \( y = f(x) \), the area between \( x = a \) and \( x = b \) is calculated using:\[ \int_{a}^{b} f(x) \, dx \]In your exercise, the function \( y = t e^{-t} \) is defined between 0 and 2 on the t-axis. Thus, by solving the definite integral \( \int_{0}^{2} t e^{-t} \, dt \), you can find the exact area under this part of the curve, which represents the total "accumulated influence" of the function in this range.Solving for this definite integral involves integrating by parts and substituting in the limits of integration to find an exact result. The final answer gives both a precise mathematical form and numerical approximation to enhance comprehension.