Evaluate Integral: sinπ(k/2) with k = 1 to 6

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In summary, the value of the given integral is 0. To evaluate the integral, the formula for the integral of sine function, which is -cos(x) + C, is used. The value of the integral is 0 because the sine function has a period of 2π and repeats itself when k=1 to 6, resulting in a net value of 0. The value of the integral cannot be negative as the sine function is always positive or zero. Evaluating this integral can be useful in applications involving periodic functions and can help in understanding the behavior of the sine function and its relationship with other functions.
  • #1
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Question,
Evaluate:
[tex]\sum_{k=1}^{\ 6}\ f(xk)[/tex]
where
[tex]\ xk\ = \ k/2[/tex]
and
[tex]\ f(x)=sin\pi\ x[/tex]

OK,
does this mean that that I should form a RiemannSum with;
[tex]sin\pi\ (k/2)* delt(k/2)= \sum_{k=1}^{\ 6}f(k/2)delt(k/2) [/tex]

Im confused.
 
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  • #2
I cannot see your problem, it is a trivial substitution

[tex]\sum_{k=1}^{\ 6}\ f(xk) = \sum_{k=1}^{\ 6}\ \sin(k\frac{\pi}{2})=1[/tex]
 
  • #3
Thankyou,
sometimes I get muddled when working with calculus, and often percieve things as being a lot harder than they are.

Thanks again.
 
  • #4
Ok, quite an usual occurrence!
 

FAQ: Evaluate Integral: sinπ(k/2) with k = 1 to 6

What is the value of the given integral?

The value of the given integral is 0.

How do you evaluate the integral?

To evaluate the integral, we need to use the formula for the integral of sine function, which is -cos(x) + C. In this case, we need to plug in the given value of k, which is 1 to 6, into the formula and then subtract the value of the integral at k=1 from the value of the integral at k=6.

Why is the value of the integral 0?

The value of the integral is 0 because the sine function has a period of 2π, and when k=1 to 6, the function repeats itself and cancels out, resulting in a net value of 0.

Can the value of the integral be negative?

No, the value of the integral cannot be negative because the sine function is always positive or zero.

What is the significance of evaluating this integral?

Evaluating this integral can be useful in applications involving periodic functions, such as in physics and engineering. It can also help in understanding the behavior of the sine function and its relationship with other functions.

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