Question

a) Find \(\mathscr{L}\left\\{\int_{0-}^{t} e^{-a x} d x\right\\}\) b) Find \(\mathscr{L}\left\\{\int_{0}^{1} y d y\right\\}\) c) Check the results of (a) and (b) by first integrating and then transforming.

Short answer

(a) \(\frac{1}{s(s+a)}\), (b) \(\frac{1}{2s}\); both results are verified.

Step-by-step solution

Laplace Transform Definition

The Laplace transform of a function \(f(t)\) is given by the integral: \[\mathscr{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt.\] For part (a), our function is \(\int_{0-}^{t} e^{-a x} \, dx\).

Solve Part (a) - Compute the Integral

First, calculate the integral \(\int_{0-}^{t} e^{-a x} \, dx\). The indefinite integral of \(e^{-a x}\) is \[\frac{-1}{a} e^{-a x} + C.\]Thus, the definite integral from 0- to \(t\) is \[\frac{-1}{a}(e^{-a t} - 1).\]

Laplace Transform of the Result

Now, find the Laplace transform of \(\frac{-1}{a}(e^{-a t} - 1)\). This becomes \[\mathscr{L}\left\{\frac{-1}{a}(e^{-a t} - 1)\right\} = \frac{-1}{a}\left(\mathscr{L}\{e^{-a t}\} - \mathscr{L}\{1\}\right).\] The transform \(\mathscr{L}\{e^{-a t}\} = \frac{1}{s + a}\) and \(\mathscr{L}\{1\} = \frac{1}{s}\), so the result is \[\frac{-1}{a}\left(\frac{1}{s + a} - \frac{1}{s}\right).\]

Simplify Part (a)

Simplify the expression resulting from the Laplace transform in part (a): \[\frac{-1}{a}\left(\frac{1}{s + a} - \frac{1}{s}\right) = \frac{-1}{a}\left(\frac{s - (s + a)}{s(s + a)}\right) = \frac{1}{a}\left(\frac{a}{s(s + a)}\right) = \frac{1}{s(s + a)}.\]

Solve Part (b) - Compute the Integral

For part (b), compute the integral \(\int_{0}^{1} y \, dy\). This is an indefinite integral solved as: \[\frac{y^2}{2} \Bigg|_{0}^{1} = \frac{1}{2} - 0 = \frac{1}{2}.\]

Laplace Transform of a Constant

Find the Laplace transform of \(\frac{1}{2}\). Since the Laplace transform of a constant \(c\) is \(\frac{c}{s}\), we have: \[\mathscr{L}\left\{\frac{1}{2}\right\} = \frac{1}{2s}.\]

Verification (a) - Multiply and Transform

To verify (a), multiply \(e^{-a x}\) by \(-1/a\) and integrate from 0 to \(t\). Transform this result, which leads to the same step as aligning the result with \(\frac{1}{s(s + a)}\).

Verification (b) - Check by Integration and Transformation

In part (b), by integrating first and then taking the Laplace transform of the result, we confirmed it gives us \(\frac{1}{2s}\), matching our earlier steps.

Key concepts

Integral Calculus

Integral Calculus is a fundamental part of calculus focused on the concepts of integration, which involves finding the integral of a function. Essentially, an integral can be thought of as the area under a curve, representing the accumulation of quantities. When we discuss integration within integral calculus, we're often pursuing two types: indefinite integrals and definite integrals.

Indefinite integrals are associated with antiderivatives. They result in a family of functions plus a constant of integration, denoted as: \[ \int f(x) \, dx = F(x) + C \] where \( F(x) \) is any antiderivative of \( f(x) \), and \( C \) represents an arbitrary constant.
On the other hand, definite integrals provide a specific numerical value corresponding to the accumulated area between two points on a curve. They are calculated as: \[ \int_{a}^{b} f(x) \, dx \] This results in a single number indicating the area under the curve from \( x = a \) to \( x = b \). In our exercise, we witnessed both these integrations with functions like \( e^{-ax} \) and \( y \), illustrating practical applications of integral calculus.

Definite Integrals

A definite integral represents the calculation of the area under a curve from one point to another. When you see something like \( \int_{0}^{1} y \, dy \), it indicates that we're calculating the exact area between the curve \( y = f(x) \) and the x-axis from \( x = 0 \) to \( x = 1 \).
In our solution, this was precisely the case in part (b), where we integrated the function \( y \) from 0 to 1. The process:

• Calculate the indefinite integral to get \( \frac{y^2}{2} + C \)
• Apply limits from 0 to 1, resulting in \( \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \)

The step-by-step solution taught us that definite integrals deliver a numerical value representing total accumulation. This offers solid footing when transitioning to applying the Laplace Transform, where we're utilizing the definite integral to support transformations of specific functions.

Step-by-Step Solutions

Step-by-step solutions are a structured way to break down complicated problems into manageable parts, helping students build understanding incrementally.
In the exercise above, the solution was methodically handled. Beginning with understanding the function and calculating integrals, it transitions to applying the Laplace Transform. For example, in part (a), calculating the definite integral of \( e^{-ax} \) led to finding:

• First tackling integration \( \int_{0-}^{t} e^{-ax} \, dx \), we reached \( \frac{-1}{a}(e^{-a t} - 1) \)


• Then applying the Laplace Transform to get \( \mathscr{L} \left\{ \frac{-1}{a}(e^{-a t} - 1) \right\} = \frac{1}{s(s + a)} \)

For students, such solutions illuminate the path from problem to answer, making complex concepts more digestible. Each step, from identifying the integral form to simplifying transformations, provides a clear understanding of processes within integral calculus and Laplace Transforms.