Finding area under cosx using midpoint rule

[pvpkillerx]

Use the midpoint rule to approximate the following integral:
∫sin(x) dx

This is what I did:
Δx = (1-0)/4 = 1/4

1/4(f(1/8) + f(3/8) + f(5/8) + f(7/8) + f(9/8))

But the answer i get is wrong. Is that the correct midpoint rule formula, and are the values I plugged in right? Any help is appreciated, thanks.

 

[LCKurtz]

Use the midpoint rule to approximate the following integral:
∫sin(x) dx

This is what I did:
Δx = (1-0)/4 = 1/4

1/4(f(1/8) + f(3/8) + f(5/8) + f(7/8) + f(9/8))

But the answer i get is wrong. Is that the correct midpoint rule formula, and are the values I plugged in right? Any help is appreciated, thanks.

Are the limits on your integral 0 and 1?. How many points in your partition? Is 9/8 in your interval?

 

[pvpkillerx]

Ye, its from 1 to 0, and n = 4.

 

[LCKurtz]

Are the limits on your integral 0 and 1?. How many points in your partition? Is 9/8 in your interval?[/QUOTE]

Ye, its from 1 to 0, and n = 4.

What about my last question?

 

[pvpkillerx]

I tried with and without 9/8, both don't give me the right answer.

 

[LCKurtz]

I tried with and without 9/8, both don't give me the right answer.

Why would you try it with 9/8 in the first place? Unless you show us your work how can we help you find what you are doing wrong? Maybe something simple like having your calculator in degree mode instead radians? Show us your calculations.

 

[pvpkillerx]

ohh, i got the answer, u were right, the calculator was suppose to be in radians mode -_- oops. Thanks.

 

[NascentOxygen]

Use the midpoint rule to approximate the following integral:
¢

 

[NascentOxygen]

Use the midpoint rule to approximate the following integral: \intsin(x) dx

But title says: Finding area under cosx using midpoint rule

Better establish whether sine or cosine