ℒ{f(t)} =∫0^1(2 t+1) e^-s t d t=.(-(2)/(s) t e^-s t-(2)/(s^2) e^-s t-(1)/(s) e^-s t)|0 ^1

step 1 Hello everyone. Here we have to evaluate d square by dx square of double integral 0 to x 0 to x

ℒ{f(t)} =∫0^1(2 t+1) e^-s t d t=.(-(2)/(s) t e^-s...

$$\begin{aligned} \mathscr{L}\{f(t)\} &=\int_{0}^{1}(2 t+1) e^{-s t} d t=\left.\left(-\frac{2}{s} t e^{-s t}-\frac{2}{s^{2}} e^{-s t}-\frac{1}{s} e^{-s t}\right)\right|_{0} ^{1} \\ &=\left(-\frac{2}{s} e^{-s}-\frac{2}{s^{2}} e^{-s}-\frac{1}{s} e^{-s}\right)-\left(0-\frac{2}{s^{2}}-\frac{1}{s}\right)=\frac{1}{s}\left(1-3 e^{-s}\right)+\frac{2}{s^{2}}\left(1-e^{-s}\right), \quad s>0 \end{aligned}$$

$$\begin{aligned}
\mathscr{L}\{f(t)\} &=\int_{0}^{1}(2 t+1) e^{-s t} d t=\left.\left(-\frac{2}{s} t e^{-s t}-\frac{2}{s^{2}} e^{-s t}-\frac{1}{s} e^{-s t}\right)\right|_{0} ^{1} \\
&=\left(-\frac{2}{s} e^{-s}-\frac{2}{s^{2}} e^{-s}-\frac{1}{s} e^{-s}\right)-\left(0-\frac{2}{s^{2}}-\frac{1}{s}\right)=\frac{1}{s}\left(1-3 e^{-s}\right)+\frac{2}{s^{2}}\left(1-e^{-s}\right), \quad s>0
\end{aligned}$$

Key Concepts

Definite Integration with Finite Limits

In some cases, rather than integrating from zero to infinity, the integral is taken over a finite interval, reflecting the support of the function being transformed. This approach is particularly important when dealing with functions that are defined or relevant on a bounded domain, and it often requires careful evaluation of the integral at the endpoints to capture the complete behavior of the transform.

Integration of Products of Functions

When evaluating Laplace transforms, especially those involving products of a polynomial term and an exponential, techniques such as integration by parts are often employed. This method helps in breaking down the integral into simpler parts, making it easier to compute the transform by successively reducing the degree of the polynomial or isolating the exponential factor.

Exponential Kernel in Integration

The use of the exponential function as a kernel in the Laplace transform serves to weigh the contributions of the original function over time. This exponential decay factor influences how different parts of the time-domain function are integrated, serving to 'dampen' or reduce the impact of function values at later times, and it is central to understanding the transform's convergence properties.

Laplace Transform

The Laplace transform is an integral transform that converts a time-domain function into a complex frequency-domain function. It is defined by the integral of the product of the function and an exponential kernel, usually over the interval from zero to infinity. This transformation is a powerful tool in solving differential equations and analyzing systems, as it simplifies differential operations into algebraic ones.

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