Carry negative parameter bounds through the substitution

C: x=−1/tx=-1/t, y=1/4−t2y=1/\sqrt{4-t^2}, −2<t<0-2<t<0. S is bounded by C, x=1/3,x=1/2x=1/\sqrt3,x=1/\sqrt2 and the xx-axis.

Show its area is ∫ab1t24−t2dt\int_a^b\frac1{t^2\sqrt{4-t^2}}dt, determining exact constants a,ba,b. Using t=2sin⁡θt=2\sin\theta, find the exact area. marks

[2011/SAJC/Prelim/II/2(b)]

Solution

(a)

Parametric area

The curve CC has parametric equations x=−1tx=-\dfrac1t, y=14−t2y=\dfrac{1}{\sqrt{4-t^2}} for −2<t<0-2<t<0. The region SS is bounded by CC, the lines x=13x=\frac1{\sqrt3} and x=12x=\frac1{\sqrt2}, and the xx-axis. Show that its area is ∫ab1t24−t2 dt\displaystyle\int_a^b\frac{1}{t^2\sqrt{4-t^2}}\,dt, stating the exact values of aa and bb.

Explore this idea

Hint

  1. Write the area in Cartesian form first: ∫y dx\int y\,dx between the two vertical lines.

  2. Convert the limits by solving x=−1tx=-\frac1t for tt. Remember tt is negative.

  3. Replace dxdx with dxdt dt\frac{dx}{dt}\,dt.

Answer · (a)

a=−3,b=−2a=-\sqrt3,b=-\sqrt2.

Explore the idea

The region under the curve between the two vertical lines

Written in Cartesian form, y = x/√(4x² − 1). The curve falls from 1 at x = 1/√3 to about 0.71 at x = 1/√2.

Number of stripsn = 4
Reset the controls

Where does one strip’s area come from?

1 of 4

Click a strip in the graph to inspect it. The darker rectangle is this strip. Its red marks use the left edge at x* = 0.57735.

upperlowerwidth Δx
Zoomed-in rectangle: width and height are scaled independently. The sample sets its upper and lower values across the whole strip.
Sample x*
0.57735
Width Δx
0.032439
Upper value
1
Lower value
0
Height: upper − lower
1
Strip area
0.032439

0.032439 ×(1 − 0)≈ 0.032439

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

Strip width
0.032
0.130 ÷ 4
Sum of strips
0.11078
Exact, ¼(1 − 1/√3)
0.10566
Error
+0.00511
4.84% too big

How the sum improves as n grows

  • C in Cartesian form
  • X-axis, the bottom of S
  • Each of width (1/√2 − 1/√3)/n and height y
  • Exact area ¼(1 − 1/√3) ≈ 0.10566
Key idea

The picture is a thin sliver of height about 0.7 to 1 and width 0.13. Its area, about 0.1060.106, is what the substitution must reproduce.

Ideas to try
  1. Compare the curve height at the two ends. Which end of the t-interval is each?
  2. How many strips do you need for 3 decimal places?
(b)

Trigonometric substitution

Using the substitution t=2sin⁡θt=2\sin\theta, find the exact area of SS.

Hint

  1. Change the limits again: what is θ\theta when t=−3t=-\sqrt3 and when t=−2t=-\sqrt2?

  2. Simplify 4−4sin⁡2θ=2cos⁡θ\sqrt{4-4\sin^2\theta}=2\cos\theta and cancel.

  3. The integrand becomes a multiple of csc⁡2θ\csc^2\theta.

Answer · (b)

A=14(1−1/3)A=\frac14(1-1/\sqrt3).

Takeaway

Convert limits twice

One substitution from xx to tt, then another from tt to θ\theta. Each needs its own limits, and with a negative parameter it pays to write the sign next to every one.