# Thread: Express the integral as a limit,

1. ## Express the integral as a limit,

Express the integral as a limit, integral from -5 to -1 of x sin (x) dx using left endpoints
so it would be i= 1 to n f(-5+ (i-1) 4/n * 4/n and then I don't know how to continue

2. ## Re: Express the integral as a limit,

Originally Posted by sic
Express the integral as a limit, integral from -5 to -1 of x sin (x) dx using left endpoints
so it would be i= 1 to n f(-5+ (i-1) 4/n * 4/n and then I don't know how to continue
looks like you've just about got it. It's easier here to use 0 as the sum starting index.

$$\sum_{i=0}^{n-1}f\left(-5+\frac{4i}{n}\right) \frac{4}{n}$$

now make use of the fact that

$$f(x)=x \sin(x)$$

and note that

$$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$$

you'll end up with an approximation formula for the integral in terms of n. Then take the limit as n goes to infinity.

3. ## Re: Express the integral as a limit,

Is it the limit as n goes to infinity of the sum i=0, n-1, (sin(-5) cos(4i/n) + cos (-5) sin(4i/n)) 4/n then?

4. ## Re: Express the integral as a limit,

Not quite.

in the sum

$$x_i=-5+\frac{4i}{n}$$

$$dx = \frac{4}{n}$$

$$f_i=x_i \sin(x_i) = \left(-5+\frac{4i}{n}\right) \sin\left(-5+\frac{4i}{n}\right)$$

so your sum becomes

$$\sum_{i=0}^{n-1}f_i dx = \sum_{i=0}^{n-1} \left(-5+\frac{4i}{n}\right) \sin\left(-5+\frac{4i}{n}\right) \frac{4}{n}$$

you can work the algebra to finish it up

a numerical check shows this does converge (very slowly) to the correct answer.