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Chapter 8 · Application of Integrals

Miscellaneous · Q3: Area bounded by y=sin⁡xy = \sin x from x=0x = 0 to x=2πx = 2\pi

Miscellaneous ExerciseA curve below the axis: split where it crosses · Miscellaneous Exercise
The question

3. Find the area bounded by the curve y=sin⁡xy = \sin x between x=0x = 0 and x=2πx = 2\pi⁠.

Exam answer

(figure) The curve y=sin⁡xy = \sin x meets the xx-axis where sin⁡x=0\sin x = 0⁠: at x=0,π,2πx = 0, \pi, 2\pi⁠.

It is above the axis on [0,π][0, \pi] and below it on [π,2π][\pi, 2\pi]⁠.

A=∫0πsin⁡x dx+∣∫π2πsin⁡x dx∣=[−cos⁡x]0π+∣[−cos⁡x]π2π∣=(−cos⁡π+cos⁡0)+∣−cos⁡2π+cos⁡π∣=(1+1)+∣−1−1∣=2+2\begin{aligned} A &= \int_0^{\pi}\sin x\,dx \\ &\quad + \left|\int_{\pi}^{2\pi}\sin x\,dx\right| \\ &= \left[-\cos x\right]_0^{\pi} \\ &\quad + \Big|\left[-\cos x\right]_{\pi}^{2\pi}\Big| \\ &= (-\cos\pi + \cos 0) \\ &\quad + \lvert -\cos 2\pi + \cos\pi\rvert \\ &= (1 + 1) + \lvert -1 - 1\rvert \\ &= 2 + 2 \end{aligned}

∴\therefore Required area =4= 4 sq. units.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Between 0 and 2π2\pi the sine curve makes one arch above the xx-axis (from 0 to π\pi⁠) and one arch below it (from π\pi to 2π2\pi⁠). The integral over the lower arch is negative, −2-2⁠, and would cancel the upper arch's +2+2 if we integrated in one go. An area is never negative, so we split at the crossing x=πx = \pi⁠, take the size of each part and add: 2+∣−2∣=42 + \lvert -2\rvert = 4⁠.

Given, to find, and the plan
Given
The curve y=sin⁡xy = \sin x⁠, the lines x=0x = 0 and x=2πx = 2\pi⁠, and (as always for "the area bounded by the curve") the xx-axis.
To find
The area of the region between the curve and the xx-axis from x=0x = 0 to x=2πx = 2\pi⁠.
Plan
(1) Find where the curve crosses the xx-axis. (2) Find where it is above and where it is below. (3) Write the area as a sum of integrals, with bars round the part below the axis. (4) Evaluate each part. (5) Add.

Basics you need

1Area below the axis

Where y<0y < 0⁠, each strip y dxy\,dx is negative, so the integral there is negative. The area is its absolute value. If the curve is partly below and partly above, A=∣A1∣+A2A = |A_1| + A_2 (Section 8.2, Remark). Basics: area below the axis

2Where sin⁡x\sin x is zero, positive and negative

sin⁡x\sin x is the height of the point at angle xx on the unit circle. It is 0 at x=0,π,2πx = 0, \pi, 2\pi⁠, positive for 0<x<π0 < x < \pi (upper half of the circle) and negative for π<x<2π\pi < x < 2\pi (lower half).

3The antiderivative of sin⁡x\sin x

∫sin⁡x dx=−cos⁡x+C\int\sin x\,dx = -\cos x + C⁠, because ddx(−cos⁡x)=sin⁡x\frac{d}{dx}(-\cos x) = \sin x (Chapter 7, Section 7.2). Values: cos⁡0=1\cos 0 = 1⁠, cos⁡π=−1\cos\pi = -1⁠, cos⁡2π=1\cos 2\pi = 1⁠. Basics: the definite integral

4Split at each crossing

∫ab=∫ac+∫cb\int_a^b = \int_a^c + \int_c^b (property P2, Chapter 7, Section 7.10), with cc where the curve crosses the axis. Lesson: below the axis, take the size

Solution

  1. Where does the curve meet the xx-axis?

    On the xx-axis, y=0y = 0⁠. So we solve sin⁡x=0\sin x = 0 for 0≤x≤2π0 \le x \le 2\pi⁠. The sine of an angle is zero exactly when the angle is a whole multiple of π\pi⁠, so the solutions are

    x=0,x=π,x=2π.x = 0, \quad x = \pi, \quad x = 2\pi .
    The crossings inside the interval are where the curve may change from above the axis to below it, so they are where the integral must be split. The ends 0 and 2π2\pi are crossings too, but they are already limits; only x=πx = \pi is a new split point.
    sin⁡x=0  ⟺  x=nπ\sin x = 0 \iff x = n\pi⁠, nn an integer. In [0,2π][0, 2\pi]⁠: n=0,1,2n = 0, 1, 2⁠.

  2. Which arch is above, which is below?

    For 0<x<π0 < x < \pi⁠, sin⁡x>0\sin x > 0⁠: the curve is above the axis, rising to its top (π2,1)\left(\frac{\pi}{2}, 1\right)⁠. For π<x<2π\pi < x < 2\pi⁠, sin⁡x<0\sin x < 0⁠: the curve is below the axis, falling to its bottom (3π2,−1)\left(\frac{3\pi}{2}, -1\right)⁠.

    x y π/2 π 3π/2 2π 1 −1 O (π/2, 1) (3π/2, −1) 2 |−2| = 2 y = sin x
    y=sin⁡xy = \sin x from 0 to 2π2\pi⁠, drawn with equal scales (one unit is the same length on both axes). The upper arch (blue) has area 2; the lower arch (rose) is below the axis, its integral is −2-2 and its area is ∣−2∣=2\lvert -2\rvert = 2⁠.
    The board answer needs this figure, and it shows the trap at once: the rose arch is the blue arch turned upside down, so its signed integral exactly cancels the blue one.
    One test point on each side: sin⁡π2=1>0\sin\frac{\pi}{2} = 1 > 0 and sin⁡3π2=−1<0\sin\frac{3\pi}{2} = -1 < 0⁠. Since sin⁡x\sin x is continuous and is not zero inside either interval, its sign does not change within them. Lesson: below the axis

  3. Write the area as a sum, with bars on the lower part

    A=∫0πsin⁡x dx+∣∫π2πsin⁡x dx∣.A = \int_0^{\pi}\sin x\,dx + \left|\int_{\pi}^{2\pi}\sin x\,dx\right| .
    This is the book's rule A=∣A1∣+A2A = |A_1| + A_2 from the Remark in Section 8.2, used with the parts in the order they occur: the part above the axis counts as it is, the part below the axis is replaced by its size. The bars change nothing for the first part, so we leave them off there.
    Basics: area below the axis

  4. The upper arch, from 0 to π\pi

    ∫0πsin⁡x dx=[−cos⁡x]0π=(−cos⁡π)−(−cos⁡0)=−(−1)+1=2.\begin{aligned} \int_0^{\pi}\sin x\,dx &= \left[-\cos x\right]_0^{\pi} \\ &= (-\cos\pi) - (-\cos 0) \\ &= -(-1) + 1 = 2 . \end{aligned}
    The antiderivative is −cos⁡x-\cos x⁠. Upper value minus lower value: −cos⁡π=1-\cos\pi = 1 and −cos⁡0=−1-\cos 0 = -1⁠, so the difference is 1−(−1)=21 - (-1) = 2⁠. A positive result, as expected for an arch above the axis.
    Basics: the definite integral is F(b) − F(a)

  5. The lower arch, from π\pi to 2π2\pi

    ∫π2πsin⁡x dx=[−cos⁡x]π2π=(−cos⁡2π)−(−cos⁡π)=−1+(−1)=−2.\begin{aligned} \int_{\pi}^{2\pi}\sin x\,dx &= \left[-\cos x\right]_{\pi}^{2\pi} \\ &= (-\cos 2\pi) - (-\cos\pi) \\ &= -1 + (-1) = -2 . \end{aligned}

    So the area of the lower arch is ∣−2∣=2\lvert -2\rvert = 2⁠.

    The integral is negative because every strip sin⁡x dx\sin x\,dx here has negative height. Its size, 2, is the area. The two arches have the same area because sin⁡(x+π)=−sin⁡x\sin(x + \pi) = -\sin x⁠: the lower arch is the upper one moved right by π\pi and flipped.
    cos⁡2π=1\cos 2\pi = 1⁠, cos⁡π=−1\cos\pi = -1⁠. Basics: take the size

  6. Add the two areas

    A=2+∣−2∣=2+2=4.A = 2 + \lvert -2\rvert = 2 + 2 = 4 .

    The required area is 4 sq. units.

    Compare the single integral ∫02πsin⁡x dx=2+(−2)=0\int_0^{2\pi}\sin x\,dx = 2 + (-2) = 0⁠: it is the net signed area, not the area. The book's Example 4 does the same for y=cos⁡xy = \cos x over [0,2π][0, 2\pi]⁠, with three parts, and also gets 4.
    Book Example 4: y = cos x from 0 to 2π
Result

The area bounded by y=sin⁡xy = \sin x and the xx-axis between x=0x = 0 and x=2πx = 2\pi is

A=∫0πsin⁡x dx+∣∫π2πsin⁡x dx∣=2+2=4 sq. units.\begin{aligned} A &= \int_0^{\pi}\sin x\,dx + \left|\int_{\pi}^{2\pi}\sin x\,dx\right| \\ &= 2 + 2 = 4\ \text{sq. units}. \end{aligned}

(Textbook answer: 4.)

Check it with numbers

A computer's numerical integration of ∣sin⁡x∣|\sin x| over [0,2π][0, 2\pi] gives 4.0000000000 ✓. Of sin⁡x\sin x itself it gives 2.0000000000 over [0,π][0, \pi]⁠, −2.0000000000-2.0000000000 over [π,2π][\pi, 2\pi]⁠, and 0.0000000000 over [0,2π][0, 2\pi]⁠, the cancellation the bars prevent.

A rough check of one arch: it fits inside the rectangle of width π\pi and height 1 (area π≈3.1416\pi \approx 3.1416⁠) and contains the triangle with corners (0,0)(0, 0)⁠, (π2,1)\left(\frac{\pi}{2}, 1\right)⁠, (π,0)(\pi, 0) (area π2≈1.5708\frac{\pi}{2} \approx 1.5708⁠). Its area 2 lies between them ✓.

Common mistakes
  1. Integrating from 0 to 2π2\pi in one go: [−cos⁡x]02π=−1+1=0\big[-\cos x\big]_0^{2\pi} = -1 + 1 = 0⁠. The lower arch cancels the upper one.
  2. Putting the bars round the whole integral: ∣∫02πsin⁡x dx∣=∣0∣=0\left|\int_0^{2\pi}\sin x\,dx\right| = |0| = 0⁠. The bars go round each part below the axis separately, after splitting.
  3. Sign slips with cos⁡π=−1\cos\pi = -1: −cos⁡π=+1-\cos\pi = +1⁠, so the first arch is 1+1=21 + 1 = 2⁠, not 0.
  4. Taking ∫sin⁡x dx=cos⁡x\int\sin x\,dx = \cos x: the antiderivative is −cos⁡x-\cos x⁠. With the wrong sign the first arch comes out as −2-2⁠.
What to take away
  1. Before integrating for an area, find where the curve crosses the xx-axis inside the interval, and split there.
  2. Take the size of each part below the axis, then add: A=∣A1∣+A2A = |A_1| + A_2⁠.
  3. One arch of sine has area 2: ∫0πsin⁡x dx=2\int_0^{\pi}\sin x\,dx = 2⁠. The same idea with a cubic is Miscellaneous Q4.

Every fact used, and where it comes from

FactUsed inSource
sin⁡x=0  ⟺  x=nπ\sin x = 0 \iff x = n\pi⁠; sign of sin⁡x\sin x on (0,π)(0, \pi) and (π,2π)(\pi, 2\pi)Steps 1 and 2Class XI, Trigonometric Functions (unit circle)
Part below the axis: take the absolute value; A=∣A1∣+A2A = |A_1| + A_2Steps 3 and 5Section 8.2, Remark (Figs 8.3 and 8.4) (Basics)
P2: ∫ab=∫ac+∫cb\int_a^b = \int_a^c + \int_c^bStep 3Chapter 7, Section 7.10 (Lesson)
∫sin⁡x dx=−cos⁡x+C\int\sin x\,dx = -\cos x + CSteps 4 and 5Chapter 7, standard formulae, Section 7.2
∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)Steps 4 and 5Chapter 7, Theorem 2, Section 7.8.3 (Basics)
The same method for y=cos⁡xy = \cos xStep 6Example 4, Miscellaneous Examples of Chapter 8 (Lesson)
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