Integral sum, splitting the interval into equal parts

**Revy** · August 23rd 2010, 02:36 AM

I need to compile and calculate the function's $f(x) = sin x$ sum of the integral in interval $[0, \pi]$ , while splitting the interval into $n = 6$ equal-length parts.

I do know how to do $\int_{0}^{\pi} sinx dx$ , but how do I do this while splitting it in intervals?

I didn't attend lectures, so I'm stuck at easies parts

can it look like this?
$\sum_{n=1}^{6}\int_{0}^{\pi} sinx dx$

**Ackbeet** · August 23rd 2010, 03:02 AM

Are you supposed to approximate the integral by splitting it up into six equal chunks? Or are you supposed to find the exact value by splitting it up into six equal chunks?

**Revy** · August 23rd 2010, 03:11 AM

Well, task doesn't mention approximation, so answer should be exact value.

**HallsofIvy** · August 23rd 2010, 03:32 AM

My first thought was that this was asking you to use a "Riemann sum" approximation to the integra.

But if you say it must be an exact value, then just do this: the interval $[0, \pi]$ has length $\pi$ and dividing it into 6 equal parts means each part is of length $\frac{\pi}{6}$ .

it would NOT be " $\sum_{n=1}^6 \int_0^\pi sin(x)dx$ ", that would just be 6 times $\int_0^\pi sin(x)dx$ .

Instead, you want $\sum_{n=1}^6 \int_{(n-1)\pi/6}^{n\pi/6} sin(x) dx$ which is just
$\int_0^{\pi/6} sin(x) dx)+ \int_{\pi/6}^{\pi/3} sin(x) dx+ \int_{\pi/3}^{\pi/2} sin(x)dx$ $+ \int_{\pi/2}^{2\pi/3} sin(x)dx+ \int_{2\pi/3}^{5\pi/5} sin9x)dx+ \int_{5\pi/6}^{\pi} sin(x) dx$ .

But I still think you should check to be sure you are not expected to do a Riemann sum instead.

**Revy** · August 23rd 2010, 05:32 AM

In program I did found 'Riemann sum', but in task neither it, nor approximation is mentioned

From your derivation I get $\cos0 - \cos\pi = 2\sin^{2}\frac{\pi}{2}$

To be safe it would be good to see it done with Reimann sum, I did try to find out how to get it from theory, but without example I still won't be able to use this formula

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