Area between two trig functions (1 Viewer)

[sinophile]

Here is the exact question from the Fitzpatrick textbook if it offers any clues:

"Find the points of intersection of the curves y=sin2x and y=cos2x for the values of x in the interval [0,pi] and calculate the measure of the area between the two curves in this interval."

Okay, so finding the points of interseciton is straightforward. But how would you find the area between the curves? You can't just integrate, because it asked for area and some sections are negative. You can't integrate with respect to y either, because you end up with .. weird shit. Refer to diagram.

 

[Prashant Sallan]

im just guessing... but find the inverse function?

 

[Drongoski]

Here is my attempt

(sorry: can't get LaTeX up)

Where 2 curves intersect: sin 2x = cos 2x ==> tan 2x = 1 ==> 2x =pi/4 & 5pi/4

==> x = pi/8 & 5pi/8 ( please sub in x values and find corresp y values)

reqd area = integral (from pi/8 to 5pi/8) of [sin 2x - cos 2x] wrt x

= (-0.5 cos 5pi/4 + 0.5 sin 5pi/4 ) - (- 0.5 cos pi/4 - 0.5 sin pi/4)

= [1/(2 sqr(2))+ 1/(2sqr(2))] - [-1/(2sqr(2)) - 1/(2sqr(2)) ]

=2[1/2sqr(2) + 1/2sqr(2) ]

= 2 / sqr(2)

= sqr(2)

What's the answer ?

Edit: sinophile - how do u get such a beautiful graph up ?

 

[Trebla]

Consider the positive area (which is half the total area) and split the area into sections

 

[sinophile]

Yeah, forgot the mention the answer was square root of 2...

 

[sinophile]

Consider the positive area (which is half the total area) and split the area into sections

I don't get it. How does this find the positive area? How can you assume that positive and 'negative' areas are of the same magnitude?

Here is my attempt

(sorry: can't get LaTeX up)

Where 2 curves intersect: sin 2x = cos 2x ==> tan 2x = 1 ==> 2x =pi/4 & 5pi/4

==> x = pi/8 & 5pi/8 ( please sub in x values and find corresp y values)

reqd area = integral (from pi/8 to 5pi/8) of [sin 2x - cos 2x] wrt x

= (-0.5 cos 5pi/4 + 0.5 sin 5pi/4 ) - (- 0.5 cos pi/4 - 0.5 sin pi/4)

= [1/(2 sqr(2))+ 1/(2sqr(2))] - [-1/(2sqr(2)) - 1/(2sqr(2)) ]

=2[1/2sqr(2) + 1/2sqr(2) ]

= 2 / sqr(2)

= sqr(2)

What's the answer ?

Edit: sinophile - how do u get such a beautiful graph up ?

I used graphmatica. Although im unsure about whether the textbook answer is correct, because it asked for area, and square root of 2 is just the solution if you integrated it. Unless im missing something here.

 

[Trebla]

I don't get it. How does this find the positive area? How can you assume that positive and 'negative' areas are of the same magnitude?

In this particular example, it's obvious that the area you want to find is symmetric. The graphs of y = sin 2x and y = cos 2x are identical in shape, they are only different in relative position. Draw the lines x = pi/8 and x = pi/4 on the diagram. Referring to the positive (y > 0) area only, the area in the interval pi/8 < x < pi/4 is the area under y = sin 2x minus area under y = cos 2x, whereas in the interval pi/4 < x < pi/2, the area is under the curve y = sin 2x only.

You should get the same answer as applying integration between two curves as normal since y = sin 2x is always above y = cos 2x in this case. The method I proposed is generally useful if you have something more complicated.

 

[sinophile]

In this particular example, it's obvious that the area you want to find is symmetric. The graphs of y = sin 2x and y = cos 2x are identical in shape, they are only different in relative position. Draw the lines x = pi/8 and x = pi/4 on the diagram. Referring to the positive (y > 0) area only, the area in the interval pi/8 < x < pi/4 is the area under y = sin 2x minus area under y = cos 2x, whereas in the interval pi/4 < x < pi/2, the area is under the curve y = sin 2x only.

You should get the same answer as applying integration between two curves as normal since y = sin 2x is always above y = cos 2x in this case. The method I proposed is generally useful if you have something more complicated.

I see. Confusing.

So would this solution be correct also?

edit: yeah it is. thanks.

 

[gurmies]

When you're considering areas between two curves, it doesn't matter whether the area is under or above x-axis. Hard for me to articulate why, but if you think about what you're actually doing when you find the area between 2 curves, it should become apparent =)

 

[Drongoski]

When you're considering areas between two curves, it doesn't matter whether the area is under or above x-axis. Hard for me to articulate why, but if you think about what you're actually doing when you find the area between 2 curves, it should become apparent =)

Be careful though. If the 2 functions intersect at more than 2 points, u must not forget to break up into distinct regions in which y1 > y2 in one, y2 > y1 next and so on; of course if ur worried about the signs and just want the total area you can take the absolute area of the definite integral of each region and not worry too much whether you use y1 - y2 or y2 - y1 as integrand.

 

[gurmies]

^ Correct.

 

[Drongoski]

Now that I've got LaTeX back, it's easier to see as: