Question

If \(\int_{1}^{5} t^{n} \delta(t-2) d t=8\), what is the exponent \(n\) ?

Short answer

Question: Given the integral of the function \(t^n \delta(t-2)\) from 1 to 5 is equal to 8, find the value of the exponent \(n\). Answer: 3

Step-by-step solution

Recall properties of the Dirac Delta function

Recall that the Dirac delta function, \(\delta(t−a)\), has the property: $$\int_{-\infty}^{\infty} f(t) \delta(t-a) dt = f(a)$$ for any continuous function \(f(t)\). In this case, our function is \(f(t) = t^n\), and the value of \(a\) is 2.

Apply the Dirac Delta function property to the integral

Applying the property of the Dirac delta function to our integral, we get: $$\int_{1}^{5} t^{n} \delta(t-2) d t = f(2) = 2^n$$ We're given that the integral is equal to 8, so we have: $$2^n = 8$$

Solve for the exponent \(n\)

We just need to solve the equation \(2^n = 8\) for \(n\). Since \(2^3 = 8\), it's clear that \(n = 3\). So, the exponent \(n\) is 3.

Key concepts

Properties of Dirac Delta

The Dirac Delta function, denoted as \( \delta(t-a) \), is an intriguing concept in mathematics and physics, particularly known for its role in integral calculus. It acts like a 'scanning tool' that • extracts • specific function values at contained points. The core property to remember is:

• \( \int_{-\infty}^{\infty} f(t) \delta(t-a) dt = f(a) \)

This formula shows how the Dirac delta function effectively reduces an integral to evaluating the function \( f(t) \) at a specific point \( a \).
This property simplifies computations significantly when involved in integrations, as it transforms complicated integrals into more manageable expressions.Here's an essential aspect: the Dirac Delta is not a typical function. It's more like a distribution, meaning it represents an infinitely peaked function at a point \( a \), giving it its unique integration property.
The Dirac Delta is used extensively in physics and engineering, particularly in the modeling of impulse events—like a sudden force or current spike—where pinpoint precision and isolation of effects are needed.

Integral Calculus

Integral calculus is a branch of mathematics focusing on the accumulation of quantities and the areas under curves. It involves techniques and formulas for solving integrals. Integrals come in two main types:

• Indefinite integrals, which calculate general forms of antiderivatives, represented without limits as \( \int f(x) dx \).
• Definite integrals compute a specific area under a curve within set boundaries, expressed as \( \int_{a}^{b} f(x) dx \).

The definite integral can represent various real-world quantities like distance, area, or volume.
When integrating a Dirac Delta function over a range including its 'peaked' point, the integral condenses to the function's value at that specific point.
This property exemplifies how integral calculus, combined with unique functions like the Dirac Delta function, can simplify complex problems, reducing them to evaluations at specific points rather than whole intervals.

Exponentiation

Exponentiation is the mathematical operation involving numbers raised to the power of another number. It's expressed as \( a^n \), where \( a \) is the base, and \( n \) is the exponent.
This operation indicates how many times the base multiplies itself. For example, \( 2^3 \) means multiplying 2 by itself two more times: \( 2 \times 2 \times 2 = 8 \).Exponentiation is fundamental in scientific calculations and equations, playing a critical role in understanding growth patterns (like population growth), compound interest, and exponential decay (in radioactive materials).
In problems where we need to solve for the exponent, like finding \( n \) in the expression \( 2^n = 8 \), it's a straightforward process of deducing how many times to multiply the base to reach a specific value.
Understanding exponentiations help simplify seemingly complicated equations, breaking them down into simple arithmetic evaluations.