Problem 20 - Applied Calculus

Question

Use integration by parts to find each integral. \(\int t e^{-0.2 t} d t\)

Step-by-Step Solution

Answer

The integral is \(-5te^{-0.2t} + 25e^{-0.2t} + C\).

Step 1: Identify Parts for Integration by Parts

For integration by parts, identify the components of the integral: let \( u = t \), therefore \( du = dt \), and let \( dv = e^{-0.2t} dt \). Next, find \( v \) by integrating \( dv \).

Step 2: Perform Integration

Integrate \( dv = e^{-0.2t} dt \) to find \( v \). The integral of \( e^{-0.2t} \) is \( \frac{-1}{0.2}e^{-0.2t} = -5e^{-0.2t} \). Therefore, \( v = -5e^{-0.2t} \).

Step 3: Apply Integration by Parts Formula

Use the integration by parts formula \( \int u \, dv = uv - \int v \, du \). Substitute the values: \( u = t \), \( v = -5e^{-0.2t} \), \( du = dt \).

Step 4: Compute \( uv \) Term

Calculate the term \( uv = t(-5e^{-0.2t}) = -5te^{-0.2t} \).

Step 5: Calculate \( \int v \, du \)

Compute \( \int v \, du = \int -5e^{-0.2t} \, dt \). This integral evaluates to \( -5 \left( \frac{-1}{0.2} e^{-0.2t} \right) = 25e^{-0.2t} \).

Step 6: Substitute Back into Formula

Substitute back into the integration by parts formula to find the integral: \( \int t e^{-0.2 t} dt = -5te^{-0.2t} + 25e^{-0.2t} + C \), where \( C \) is the constant of integration.

Key Concepts

Definite Integrals

Definite integrals give us the area under a curve over a certain interval, say from point \(a\) to point \(b\). Let's say in a practical sense, you want to find the total distance traveled by a car given its speed over time. This is where definite integrals come into play. They provide a precise answer to the accumulated value over a range.

• Definite integrals have upper and lower limits, written as \( \int_{a}^{b} f(x) \, dx \).
• Evaluating a definite integral results in a number, which represents the area, unlike indefinite integrals that behave more like solutions or equations.
• The process often involves finding the indefinite integral first, then applying the limits of integration to that solution.

To calculate a definite integral, we first find the indefinite integral, then plug the upper limit into this, subtract the result of plugging the lower limit. This part is delicate, requiring careful calculation!
In the exercise above, using the indefinite integral we found, we can compute definite integrals by substituting the limits into the expression \(-5te^{-0.2t} + 25e^{-0.2t} + C\) and observing the results.

Indefinite Integrals

Indefinite integrals concern functions without specific limits. They represent all possible antiderivatives of a function and include an additive constant \( C \), which accounts for vertical shifts on the graph of the function.

• These do not deliver an "accumulation quantity" but rather provide general solutions as functions.
• They are expressed as \( \int f(x) \, dx = F(x) + C \), where \( F(x) \) is the antiderivative of \( f(x) \).
• Integration by parts is a technique used to solve more complex integrals, involving dividing the integral into parts that are easier to manage.

In our original exercise, integration by parts was applied because the integral \( \int t e^{-0.2t} \) couldn't be solved directly with basic integration rules. By identifying the parts, \( u = t \) and \( dv = e^{-0.2t} dt \), we transformed it into an easier form. The solution is majorly the first step in solving definite integrals or applied in varied areas like physics for more intricate calculations.

Calculus Problem-Solving

Solving calculus problems, especially involving integration, requires understanding various techniques such as substitution, integration by parts, and partial fraction decomposition. These techniques allow us to tackle a wide variety of integrals by tailoring our approach to the specific problem at hand.

• Identify the type of integral: check if it's easy, involves substitution, or needs integration by parts.
• Integration by parts is similar to the product rule for differentiation: \( \int u \, dv = uv - \int v \, du \).
• Organize and simplify the components once the technique is chosen, aiming to simplify the problem.

In integration by parts, as in the exercise, choosing \( u \) and \( dv \) correctly is critical. Here, letting \( u \) be \( t \) and \( dv \) be \( e^{-0.2t}dt \) was strategic, as the derivative of \( t \) is simple, and the integral of \( e^{-0.2t} \) is straightforward with an initial factor adjustment. Successful integration often comes down to selecting the right strategy and being meticulous in your calculations. This approach to problem-solving in calculus builds confidence and depth in dealing with tougher problems.