∫ {sin(x) - cos(x)} dx : how do I calculate area between curves?

Question

2π

∫ sin(x) - cos(x) dx

0

How do I find the absolute value of the area between the curves?

I get each time zero!

Even if I try to find the area of one curve only, the area above and below the x-axis cancels out, thus result in zero.

Please show the steps, especially the part where the area is calculated

Answer 1

That's the correct answer. It does cancel out!

Answer 2

Your doubt is absolutely sensible.

As you correctly pointed out, you have to find out:

2π

∫ | sinx - cosx | dx

0

and, to get rid of the abs value, you have to take into account the following cases:

| sinx - cosx | = sinx - cosx if sinx ≥ cosx

| sinx - cosx | = cosx - sinx if sinx ≤ cosx

thus, as you can verify looking at sine and cosine function graphs,

and x ranging from 0 to 2π:

sinx ≥ cosx over the interval [π/4, 5π/4]

instead,

sinx ≤ cosx over the intervals [0, π/4] and [5π/4, 2π]

hence you have to split your integral into:

2π

∫ | sinx - cosx | dx =

0  

∫ [0 to (π/4)] (cosx - sinx) dx + ∫ [(π/4) to (5π/4)](sinx - cosx) dx +

∫ [(5π/4) to 2π] (cosx - sinx) dx

the antiderivatives being, obviously:

∫ (sinx - cosx) dx = - cosx - sinx + C

∫ (cosx - sinx) dx = sinx + cosx + C

thus, evaluating the required area, you get:

{[sin(π/4) + cos(π/4)] - [sin(0) + cos(0)]} + {[- cos(5π/4) - sin(5π/4)] -

[- cos(π/4) - sin(π/4)]} + {[sin(2π) + cos(2π)] - [sin(5π/4) + cos(5π/4)]} =

sin(π/4) + cos(π/4) - sin(0) - cos(0) - cos(5π/4) - sin(5π/4) +

cos(π/4) sin(π/4) + sin(2π) + cos(2π) - sin(5π/4) - cos(5π/4) =

adding the similar terms,

2sin(π/4) + 2cos(π/4) - 2sin(5π/4) - 2cos(5π/4) - sin(0) - cos(0) +

sin(2π) + cos(2π) =

2 [(√2)/2] + 2 [(√2)/2] - 2 [- (√2)/2] - 2 [- (√2)/2] - 0 - 1 + 0 + 1 =

√2 +√2 + 2 [(√2)/2] + 2 [(√2)/2] =

√2 +√2 + √2 + √2 = 4√2

I hope ths helps...

Bye and good luck!

Answer 3

well, recall that in calculus we can split up an integral like this so we have the integral[sin(x)] - integral[cos(x)], well we take the integral[sin(x)]over 0..2pi, we get the -cosine function and we find that its zero, lets look at teh graph, we see that there is equal parts above and below the x axis.  so the area under that curve is zero, same for our second integral, we get the sine function, but again, equal parts above and below when we look at x=0..2pi, so that area is zero.

now add (or subtract) the 2 parts, we still get zero.  so you did it right :)

Answer 4

it is 0, if ur confused plot the equations, visual picture helps in these cases