Use identities for the integral and geometry for the regions

Graphs y=e²ˣ and y=(x+1)², intersection (−1.2785,0.07754), common y-intercept 1Source-given figure redrawn with D3. Graphs y=e²ˣ and y=(x+1)², intersection (−1.2785,0.07754), common y-intercept 1. Only labels supplied in the original question are included.xyO(−1.2785, 0.07754)1y=(x+1)²y=e²ˣ
Question diagram 1
Graphs y=e²ˣ and y=(x+1)², intersection (−1.2785,0.07754), common y-intercept 1Source-given figure redrawn with D3. Graphs y=e²ˣ and y=(x+1)², intersection (−1.2785,0.07754), common y-intercept 1. Only labels supplied in the original question are included.xyO(−1.2785, 0.07754)1y=(x+1)²y=e²ˣ

Given exponential and quadratic graphs with printed intersection and common intercept.

Curve C and line L enclosing shaded S; labelled intersection (−2,−1)Source-given figure redrawn with D3. Curve C and line L enclosing shaded S; labelled intersection (−2,−1). Only labels supplied in the original question are included.xyOCSL(−2, −1)
Question diagram 2
Curve C and line L enclosing shaded S; labelled intersection (−2,−1)Source-given figure redrawn with D3. Curve C and line L enclosing shaded S; labelled intersection (−2,−1). Only labels supplied in the original question are included.xyOCSL(−2, −1)

Given curve C and line L enclosing shaded region S, with printed intersection.

(a) Find ∫(cot⁡62x+cot⁡42x−sin⁡(x/2)sin⁡(3x/2))dx\int(\cot^62x+\cot^42x-\sin(x/2)\sin(3x/2))dx. marks

(b) The diagram shows y=e2xy=e^{2x} and y=(x+1)2y=(x+1)^2, intersection (−1.2785,0.07754)(-1.2785,0.07754) and yy-intercept 1. R is the finite region bounded by both curves. Find the volume when R is rotated 2π2\pi about the yy-axis, to 4 decimal places. marks

(c) C: x=2(sin⁡(t/2)−cos⁡(t/2))x=2(\sin(t/2)-\cos(t/2)), y=sin⁡(t/2)+cos⁡(t/2)y=\sin(t/2)+\cos(t/2), −3π/2≤t≤−π/2-3\pi/2\le t\le-\pi/2. Line L: y=−2x−5y=-2x-5, meets C at (−2,−1)(-2,-1). S is enclosed by L,C and the xx-axis as shown.

(i) Show C’s xx-intercept is −22-2\sqrt2. marks

(ii) Find the exact area of S. marks

[2020/ASRJC/Prelim/II/4]

Total marks: 14

Solution

(a)

Trigonometric integration

Find ∫(cot⁡62x+cot⁡42x−sin⁡x2sin⁡3x2)dx\displaystyle\int\left(\cot^6 2x+\cot^4 2x-\sin\tfrac x2\sin\tfrac{3x}2\right)dx.

Hint

  1. Group cot⁡6+cot⁡4\cot^6+\cot^4 and factor. Is there an identity that links cot⁡2\cot^2 and csc⁡2\csc^2?

  2. Turn the product sin⁡sin⁡\sin\sin into a difference of cosines.

  3. Integrate each part: one is a reverse chain rule.

Answer · (a)

−cot⁡5(2x)/10+sin⁡2x/4−sin⁡x/2+C-\cot^5(2x)/10+\sin2x/4-\sin x/2+C.

On intervals with sin2x≠0.

(b)

Volume of revolution

The curves y=e2xy=e^{2x} and y=(x+1)2y=(x+1)^2 intersect at (−1.2785, 0.07754)(-1.2785,\,0.07754) and cross the yy-axis at (0,1)(0,1). The region RR is the finite region bounded by the two curves. Find the volume when RR is rotated through 2π2\pi radians about the yy-axis, giving your answer to 4 decimal places.

Explore this idea

Hint

  1. RR lies to the left of the yy-axis. At each height yy, which curves bound it?

  2. Below the exact intersection height (about 0.07754), the left edge is the parabola's left branch. Above it, the left edge is the exponential.

  3. The right edge is the parabola's right branch throughout. Use washers: (outer)² − (inner)².

Answer · (b)

V=0.5593V=0.5593 cubic units (4 d.p.).

Explore the idea

A thin pot with a curved wall

R sits to the left of the y-axis; reflecting it gives the same solid and keeps the picture on the right. The outer wall changes curve at the intersection point.

Open the interactive view to explore this geometry in 3D.

(c)i

Parametric curves

A curve CC has parametric equations x=2(sin⁡t2−cos⁡t2)x=2\left(\sin\frac t2-\cos\frac t2\right), y=sin⁡t2+cos⁡t2y=\sin\frac t2+\cos\frac t2 for −3π2≤t≤−π2-\frac{3\pi}2\le t\le-\frac\pi2. Show that the xx-intercept of CC is −22-2\sqrt2.

Hint

  1. Set y=0y=0 and solve for tt in the given range.

  2. Substitute that tt into xx.

Answer · (c)i

x=−22x=-2\sqrt2.

(c)ii

Parametric area

The line LL with equation y=−2x−5y=-2x-5 meets CC at (−2,−1)(-2,-1). The region SS is enclosed by LL, CC and the xx-axis. Find the exact area of SS.

Explore this idea

Hint

  1. SS lies below the xx-axis. Is the signed integral going to be negative?

  2. Area = (area between CC and the axis) minus a small triangle cut off by LL.

  3. Convert ∫y dx\int y\,dx to tt and simplify with sin⁡2+cos⁡2=1\sin^2+\cos^2=1 and 2sin⁡cos⁡=sin⁡2\sin\cos=\sin.

Answer · (c)ii

A=π/2−5/4A=\pi/2-5/4.

Explore the idea

A sliver below the axis, cut off by the line

C is an arc of the ellipse x²/8 + y²/2 = 1. Between x = −2√2 and x = −5/2 the top of S is the axis; from −5/2 to −2 the top is the line L.

Number of stripsn = 4
Try an example

Where does one strip’s area come from?

1 of 4

The curves meet at this sample, so the strip has zero height and contributes zero. Its red marks use the left edge at x* = -2.8284.

zero heightupper = lowerwidth Δx
Zoomed-in rectangle: width and height are scaled independently. The sample sets its upper and lower values across the whole strip.
Sample x*
-2.8284
Width Δx
0.20711
Upper value
0
Lower value
0
Height: upper − lower
0
Strip area
0

0.20711 ×(0 − 0)≈ 0

Add this contribution to those of the other 3 strips to get the sum below. Displayed values are rounded.

Strip width
0.207
0.828 ÷ 4
Sum of strips
0.28893
Exact, π/2 − 5/4
0.32080
Error
−0.03187
9.94% too small

How the sum improves as n grows

  • C, part of the ellipse
  • Top of S: the x-axis, then the line L from (−5/2, 0) down to (−2, −1)
  • Height = top − bottom
  • Exact area π/2 − 5/4 ≈ 0.3208
Key idea

The parametric integral reduces to ∫(1+sin⁡t) dt\int(1+\sin t)\,dt, which gives π2−1\frac\pi2-1 for the area under the arc. Subtracting the triangle gives π2−54≈0.321\frac\pi2-\frac54\approx0.321.

Ideas to try
  1. Find the kink at x = −5/2 where the top switches from the axis to the line.
  2. Compare midpoints with n = 10.