Why Does \int^{1}_{0}sin(\pi x)sin(n\pi x)dx = 1 for n = 1 and 0 for n >= 2?

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In summary, the integral &int;<sup>1</sup><sub>0</sub> sin(&pi;x)sin(n&pi;x)dx represents the overlap between two periodic functions, sin(&pi;x) and sin(n&pi;x) over the interval [0,1]. For n=1, the overlap is perfect, resulting in an integral value of 1. For n&ge;2, the overlap is imperfect, resulting in an integral value of 0. The value of n affects the period of the function sin(n&pi;x), and as n approaches infinity, the integral value approaches 0. This integral has applications in physics, signal processing, and digital signal processing.
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leopard
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Can somebody explain to me why
[tex]\int^{1}_{0}sin( \pi x) sin (n \pi x) dx[/tex] equals 1 if n = 1 and 0 for n >= 2? (n is an integer)
 
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Just to give a hint: trigonometric addition formulas...
If this isn't sufficient to solve it, ask again.
 

What is the significance of the integral 10 sin(πx)sin(nπx)dx for n = 1 and n ≥ 2?

The integral 10 sin(πx)sin(nπx)dx represents the overlap between two periodic functions, sin(πx) and sin(nπx), over the interval [0,1]. For n = 1, the two functions have the same period and align perfectly over the interval, resulting in an overlap of 1. For n ≥ 2, the two functions have different periods, causing them to not align perfectly over the interval and resulting in an overlap of 0.

Why does the integral 10 sin(πx)sin(nπx)dx equal 1 for n = 1 and 0 for n ≥ 2?

The value of the integral is determined by the overlap between the two functions, sin(πx) and sin(nπx), over the interval [0,1]. For n = 1, the two functions have a perfect overlap, resulting in an integral value of 1. For n ≥ 2, the overlap is no longer perfect, resulting in an integral value of 0. This is due to the different periods of the two functions.

How does the value of n affect the integral 10 sin(πx)sin(nπx)dx?

The value of n determines the period of the function sin(nπx). For n = 1, the period is 2, meaning the function will complete one full oscillation over the interval [0,1]. For n ≥ 2, the period is reduced, causing the function to complete multiple oscillations over the interval. This change in period results in a different overlap between the two functions, leading to a different integral value.

What happens to the integral 10 sin(πx)sin(nπx)dx as n approaches infinity?

As n approaches infinity, the period of the function sin(nπx) decreases and the number of oscillations over the interval [0,1] increases. This results in a smaller and smaller overlap between the two functions, causing the integral value to approach 0. In other words, as n gets larger, the two functions become more and more out of phase, resulting in a smaller overlap and a smaller integral value.

Is there a specific application or real-world example where this integral 10 sin(πx)sin(nπx)dx is used?

This integral has applications in physics, specifically in the study of waves and oscillations. For example, it can be used to calculate the interference pattern of two waves with different frequencies. It also has applications in signal processing and digital signal processing, where it can be used to analyze the frequency components of a signal.

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