Definite integration (area between curves)

[meanmachine]

Homework Statement

Calculate the area between the curves y = sin 2x and y = cos x from x = to x = 90 degrees

Homework Equations

Using the difference function: ∫(upper = 90) (lower = 0) [sin2x - cos x]

The Attempt at a Solution

(sin 2x) - (cos x)

= ∫ sin 2x - cos x
= ∫-0.5 cos 2x - sin x
= [(-0.5 cos 2 x 90) - (sin 90)] - [(-0.5 cos 2 x 0) - (sin 0)
= (0.5) - (-0.5)
= 1

The answer is 1/2

 

[Mentallic]

I was sitting there scratching my head for a bit between these two lines, trying to figure out what you had done:

= ∫ sin 2x - cos x
= ∫-0.5 cos 2x - sin x

You already took the integral on the second line so don't put in the integral sign again.

= [(-0.5 cos 2 x 90) - (sin 90)] - [(-0.5 cos 2 x 0) - (sin 0)
= (0.5) - (-0.5)
= 1

The answer is 1/2

-0.5cos(180^o) = 0.5
sin(90^o) = 1

By the way, I'm getting an answer of 0.

 

[HallsofIvy]

By the way, you should understand that the calculus properties of sine and cosine, specifically that $d sin(x)/dx= cos(x)$, $d cos(x)/dx= -sin(x)$,$\int sin(x) dx= -cos(x)+ C$, and $\int cos(x)= sin(x)+ C$ are true only if x is in radians.

Here, because you are taking sine and cosine of the limits of integration, it doesn't matter whether they are in degrees or radians but if you had something like, say,
[tex]\int x sin(x) dx[/itex]
from 0 to 90 degrees, where we have an "x" outside the trig function, you would have to convert to radians to get the correct answer.