Problem 23 - Applied Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understand the Function and Limits

The function given is \( y = t e^{-t} \). We are to find the area under this curve from \( t = 0 \) to \( t = 2 \). This will be done by integrating the function over the interval \([0, 2]\).

Step 2: Setup the Integral

To find the area under the curve \( y = t e^{-t} \) from \( t = 0 \) to \( t = 2 \), we set up the definite integral:\[\int_{0}^{2} t e^{-t} \, dt\]

Step 3: Integration by Parts

We use integration by parts, where \( u = t \) and \( dv = e^{-t} \, dt \). Then \( du = dt \) and \( v = -e^{-t} \). The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\] Substituting, we get:\[- t e^{-t} \Big|_{0}^{2} + \int_{0}^{2} e^{-t} \, dt\]

Step 4: Evaluate the First Term

Evaluate \(- t e^{-t} \) from 0 to 2:\[- t e^{-t} \Big|_{0}^{2} = - (2 e^{-2}) + (0 e^{0}) = -\frac{2}{e^2}\]

Step 5: Compute the Second Integral

Compute \( \int_{0}^{2} e^{-t} \, dt \). The antiderivative is \(-e^{-t}\):\[- e^{-t} \Big|_{0}^{2} = -e^{-2} - (-e^{0}) = 1 - \frac{1}{e^2}\]

Step 6: Final Calculation

Combine the results from steps 4 and 5 to calculate the total area:\[-\frac{2}{e^2} + \left(1 - \frac{1}{e^2}\right) = 1 - \frac{3}{e^2}\]

Step 7: Solution

Thus, the exact area under the curve \( y = t e^{-t} \) from \( t = 0 \) to \( t = 2 \) is \( 1 - \frac{3}{e^2} \).

Key Concepts

Integration by Parts

Integration by parts is a technique derived from the product rule for differentiation. It's handy when you need to integrate a product of two functions. In our problem, the function is the product of two simpler functions: a polynomial and an exponential function. The rule itself is:\[ \int u \, dv = uv - \int v \, du \]Here's how it works. You choose which part of the function becomes \(u\) and which becomes \(dv\). In this exercise, we let \(u = t\) because differentiating \(t\) makes it simpler \(du = dt\), and \(dv = e^{-t} \, dt\) because its integral, \(v = -e^{-t}\), is straightforward. Once identified, apply the formula: calculate \(uv\) and replace the second integral with \(v \, du\). This transforms a complex integral into simpler ones that are more manageable.

Exponential Functions

An exponential function is one where a constant base is raised to a variable exponent. The most common base is Euler's number, denoted as \(e\), an irrational number approximately equal to 2.718. This_function_makes exponential changes that occur naturally in processes like radioactive decay or population growth. In this exercise, we encounter \(e^{-t}\), which exemplifies a decay function.

• The exponential part, \(e^{-t}\), decreases as \(t\) rises, which affects how the function behaves over the interval.
• When integrating exponential functions, they often retain their form, making them easier to work with compared to other types.

In definite integration, understanding the exponential’s effect on functions, particularly in terms of shrinkage as \(t\) increases, helps when calculating areas under curves.

Area Under a Curve

Finding the area under a curve involves integrating a function over a specific interval. This gives us a single value representing the total area, captured between the curve and the x-axis.

• The process begins by setting up a definite integral with limits, from the lower limit to the upper limit—in this case, from \(t = 0\) to \(t = 2\).
• The function \(y = t e^{-t}\) gives us a curve, and by taking the integral, we determine the space it occupies on a graph over that interval.

In this question, integration by parts was used to handle the product \(t e^{-t}\), enabling an accurate calculation of the area from \(t=0\) to \(t=2\). The final result, \( 1 - \frac{3}{e^2} \), reveals how exponential decay limits the area as \(t\) grows.