Year 12s: what will be the first way you research your future options? >>

FP3 Rectangular approximation to series

So I have to find the upper and lowers bounds for the sum of sinr where r is in degrees between 1 and 89, leaving the bounds to 4.s.f

I did this but evidently that is incorrect ...so how do I do this?

Reply 1

TeeEm

Original post by Mathematicus65

So I have to find the upper and lowers bounds for the sum of sinr where r is in degrees between 1 and 89, leaving the bounds to 4.s.f

I did this but evidently that is incorrect ...so how do I do this?

firstly draw a picture
secondly do the integration in radians

Reply 2

Mathematicus65

Original post by TeeEm

firstly draw a picture
secondly do the integration in radians

I've sketched it, and evidently sinx is an increasing function between x =1 degree and x =89 degree.

And I integrated in radians also and still got the incorrect answer, but the question says that x is specifically to be in degrees?

Reply 3

the bear

you seem to have changed the limits from "1 to 89" to "0 to 89" near the start

Reply 4

ghostwalker

Original post by Mathematicus65

And I integrated in radians also and still got the incorrect answer, but the question says that x is specifically to be in degrees?

\int \sin x \; dx= - \cos x +c

is only true if you work in radians.

So, if x is in degrees, you have to convert it to radians? What's your multiplying factor?

(edited 8 years ago)

Reply 5

DFranklin

Mathematcius65

Note that your conversion to radians needs to happen inside the integral, not by changing the limits. The limits need to match the limits for the sum, so they are actually correct as they stand.

Reply 6

Mathematicus65

Original post by DFranklin

Note that your conversion to radians needs to happen inside the integral, not by changing the limits. The limits need to match the limits for the sum, so they are actually correct as they stand.

How do you mean inside?

Reply 7

DFranklin

Original post by Mathematicus65

How do you mean inside?

The correct integral is

\int_0^{89} \sin(x^o) \,dx

That is, the issue is that in normal calculus, sin x means the sin of x measured in radians. But here x is measured in degrees. So you need to scale x from radians to degrees.

K
(Note this is a bit vague because I don't want to just give you the answer).