Understanding the Dirac Delta Function

The problem asks us to evaluate a definite integral involving the Dirac delta function. The Dirac delta function, denoted as \(\delta(t-t_0)\), is a generalized function (or distribution) that is zero everywhere except at \(t = t_0\), where it has an infinitely sharp peak. Its integral over all space is defined to be 1.

A crucial property of the Dirac delta function for evaluating integrals is its sifting property, which states:

• If \(t_0\) is within the interval of integration \([a, b]\) (i.e., \(a < t_0 < b\)), then:

  \(\mathop \smallint \nolimits_a^b f(t)\delta(t-t_0)dt = f(t_0)\)

• If \(t_0\) is outside the interval of integration \([a, b]\) (i.e., \(t_0 < a\) or \(t_0 > b\)), then:

  \(\mathop \smallint \nolimits_a^b f(t)\delta(t-t_0)dt = 0\)

Integral Evaluation Steps

Let's analyze the given integral: \(\mathop \smallint \nolimits_{ - 7}^2 \left( {{t^2} + {t^3} + 1} \right)\delta \left( {t - 3} \right)dt\)

Here, we can identify the following components:

• The function \(f(t)\) is \(t^2 + t^3 + 1\).
• The point \(t_0\) at which the delta function is centered is \(3\).
• The lower limit of integration \(a\) is \(-7\).
• The upper limit of integration \(b\) is \(2\).

Checking the Delta Function's Position

The most important step is to determine if the point \(t_0 = 3\) lies within the integration interval \([-7, 2]\). We compare \(t_0\) with the limits of integration:

• Is \(-7 < 3 < 2\)?

Clearly, \(3\) is not less than \(2\). In fact, \(3\) is greater than the upper limit of integration \(2\).

Since \(t_0 = 3\) is outside the integration interval \([-7, 2]\), according to the property of the Dirac delta function, the integral evaluates to \(0\).

Final Calculation

Given that \(t_0 = 3\) is outside the interval \([-7, 2]\), the integral becomes:

\(\mathop \smallint \nolimits_{ - 7}^2 \left( {{t^2} + {t^3} + 1} \right)\delta \left( {t - 3} \right)dt = 0\)

Therefore, the value of the integral is \(0\).