Working Rule for finding the Area

(iii) If the curve is symmetrical about x axis, or y axis, or both, then calculate the area of one symmetrical part and multiply it by the number of symmetrical parts to get the whole area.

Illustration 15  Find the area between the curves y = x² + x –2 and y = 2x, for which |x²+ x –2| + | 2x | = |x² + 3x –2| is satisfied.

Illustration 16  Find out the area enclosed by circle |z| = 2, parabola y = x² + x + 1, the curve y = [sin² x/4 + cos x/4] and x-axis (when [ . ] is the greatest integer function).

Solution:

For x ∈ [-2, 2]

=> 1 < sin² x/4 + cos x/4 < 2

∴ [sin² x/4 + cos x/4] = 1

Now we have to find out the area enclosed by the circle |z| = 2,

parabola (y - 3/4) = (x+1/2)²,

line y = 1and x-axis.

∴ Required area is shaded area in the figure.

Hence required area = √3 × 1 + (√3 – 1) +  ∫⁰₁ (x² + x + 1).dx + 2  ∫²_√3√4–x² dx 

= (2π/3 + √3 – 1/6) sq. units

Illustration 17:  Let f(x) = Max. {sinx, cos x, 1/2} then determine the area of the region bounded by the curves y = f(x), x-axis, y-axis and x = 2π.

Solution:

f(x) = Max {sin x, cos x, 1/2}

interval value of f(x)

for 0 < x < π/4, cos x

for  π/4 < x < 5π/6,  sin x

for 5π/6 < x < 5π/3, 1/2 

for   5π/3 < x < 2π       cos x

Hence required area

= ∫^π/4₀ cosx dx + ∫^5π/6π/4 sin x dx + ∫^5π/35π/6 1/2.dx + ∫^2π5π/3 cosx dx

= (5π/12 + √2 + √3) sq units.
 

Illustration 18: Find area bounded by y = |4 – x²|/4 and y = 7 - |x|

Solution: 

y = |4 – x²|/4        (I)

y = 7 – |x|       (II)

Rewriting (I) and (II)

                                        = 2 [1/2 (7 + 3)4 - ∫²₀ (4–x²/4) dx – ∫⁴₂ x²–4/4 dx] 

                                        = 2 18 + 2/3 – 16/3 + 4 + 2/3 – 2

                                        = 32 sq unit.
 

Illustration 19: Let f be a real valued function satisfying f(x/y) and lim_x–>0 f(1+x)/x = 3. Find the area bounded by the curve y = f(x), the y axis and the line y = 3

Solution: 

Given = f(x/y) = f(x) – f(y)                                                                        .... (1)

Putting x = y = 1, we get f(1) = 0

Now,    f’(x) = lim_h–>0 f(x+h) – f(x) / h

= Lim_h–>0 f(1+h/x)/h         (from(1))

= Lim_h–>0 f(1+h/x) (h/x)x

=>   f'(x) = 3/x  {since Lim_x–>0 f(1+x0/x = 3}

=>   f(x) = 3 ln x + c

Putting x = 1

=>   c = 0

=>   f(x) = 3 ln x = y (say)

Required area = ∫³_–∞ x dy = ∫³_–∞ e^y/3 dy  = 3[e^y/3]³_–∞

= 3 (e – 0) = 3e sq. units

Illustration 20:   Let A_n be the area bounded by y = tanⁿ x, x =0, y = 0 and x = p/4. Prove that for n ³ 2.   (i) A_n+A_n-2 = 1/n–1  (ii) 1/2(n+1) < A_n < 1/2(n–1)

 (ii) Obviously A_n+2< A_n< A_n-2    (from figure).

Thus 2 A_n = A_n + A_n < A_n + A_n-2 = 1/n–1 (by part (i)

Thus A_n < 1/2(n–1)              . . . . (1)

Also 2 A_n = A_n+ A_n > A_n + A_n+2 = 1/n+1, replacing n by n+2 in (i)

A_n > 1/2(n+1)                      . . . . (2)

From (1) and (2), we get 1/2(n+1) < A_n < 1/(2n–1).

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