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Q. 1. Find the area bounded by the curve x^{2} = 4y and the straight line x = 4y – 2. (1981  4 Marks)
Ans.
Solution. To find the area bounded by
x^{2} = 4y ..(1)
which is an upward parabola with vertex at (0, 0).
which is a st. line with its intercepts as –2 and 1/2 on axes. For Pt’s of intersection of (1) and (2) putting value of 4y fom (2) in (1) we get
x^{2} = x+2 ⇒ x^{2} x 2 = 0 ⇒ (x2)(x+1) = 0
⇒ x = 2,1 ⇒ y = 1,1 /4
∴ A(1,1 / 4)B(2,1).
Shaded region in the fig is the req area.
Q. 2. Show that : (1981  2 Marks)
Solution. We know that in integration as a limit sum
Similarly the given series can be written as
Q. 3. (1982  2 Marks)
Solution. … (1)
Hence Proved.
Q. 4. (1982  3 Marks)
Ans.
Solution.
Q. 5. is a point on the hyperbola x^{2} – y^{2} = 1. Show that the area bounded by this hyperbola and the lines joining its centre to the points corresponding to t_{1} and –t_{1} is t_{1}. (1982  3 Marks)
Solution. Let P(t_{1}) and Q(–t_{1}) be two points on the hyperbola.
∴ The required area = Ar (ΔLPQ) Ar ( PRQOP)
Q. 6. (1983  3 Marks)
Ans.
Solution.
Let sin x  cox x = t ⇒ as x → 0,t →1 as x → π / 4,t → 0
⇒ (cos x + sin x)dx= dt
Q. 7. Find the area bounded by the xaxis, part of the curve and the ordinates at x = 2 and x = 4. If the ordinate at x = a divides the area into two equal parts, find a. (1983  3 Marks)
Ans. a = 2√2
Solution.
If x = 4a bisects the area then we have
Q. 8. Evaluate the following (1984  2 Marks)
Ans.
Solution.
Put x = sinθ ⇒ dx =cosθ d θ
Also when x = 0, θ = 0
and when x = 1/2, θ = θ / 6
Intergrating the above by parts, we get
Q. 9. Find the area of the region bounded by the xaxis and the curves defined by (1984  4 Marks)
Ans.
Solution. To find the area bold by x  axis and curves
y = tan x, π/3 < x < π /3 … (1)
and y = cot x, π / 6 < x < 3π /2 … (2)
The curves intersect at P, where tan x = cot x, which is satisfied at x = π /4 within the given domain of x.
The required area is shaded area
Q. 10. Given a function f(x) such that (1984  4 Marks)
(i) it is integrable over every interval on the real line and
(ii) f (t + x) = f (x), for every x and a real t, then show that the integral is independent of a.
Solution.
This shows that I is independent of a.
Q. 11. Evaluate the following (1985  2½ Marks)
Ans.
Solution.
Q. 12. Sketch the region bounded by the curves and y = x  1  and find its area. (1985  5 Marks)
Ans.
Solution. The given curves are
y =x 1 … (2)
We can clearly see that (on squaring both sides of (1)) eq. (1) represents a circle. But as y is + ve sq. root, ∴ (1) represents upper half of circle with centre (0, 0 ) and radius √5.
Eq. (2) represents the curve
Graph of these curves are as shown in figure with point of intersection of and of
The required area = Shaded area
Q. 13. (1986  2½ Marks)
Ans.
Solution.
...(1)
...(2)
Adding (1) and (2), we get
Q. 14. Find the area bounded by th e curves, x^{2} + y^{2} = 25, 4y =  4 – x^{2}  and x = 0 above the xaxis. (1987  6 Marks)
Ans.
Solution. We have to find the area bounded by the curves
x^{2} +y^{2} = 25 ...(1)
...(2)
x = 0 ...(3)
and above xaxis.
Thus we have three curves
or (y  2)^{2} = 5^{2} ∴ y  2 = ±5 y = 7,y =3
y = – 3, 7 are rejected since.
y = – 3 is below xaxis and
y = 7 gives imaginary value of x. So, (I) and (II) do not intersect but II intersects xaxis at (2, 0) and (–2, 0). (I) and (III) intersect at
y = – 7 is rejected, y = 3 gives the points above xaxis. When y = 3, x = ±4. Hence the points of intersection of (I) and (III) are (4, 3) and (– 4, 3). Thus we have the shape of the curve as given in figure
.
Required area is
Q. 15. Find the area of the region bounded by the curve C : y = tan x, tangent drawn to C at x = π/4 and the xaxis. (1988  5 Marks)
Ans.
Solution. The given curve is y = tan x ...(1)
Let P be the point on (1) where x = π /4
∴ y = tan p / 4=1 i.e. coordinates of P are (π / 4,1)
∴ Equation of tangent at P is y  1 = 2(x π / 4)
or y = 2x + 1 π/2 ...(2)
The graph of (1) and (2) are as shown in the figure.
Tangent (2) meets xaxis at,
Now the required area = shaded area
= Area OPMO  Ar (ΔPLM)
Q. 16. (1988  5 Marks)
Ans.
Solution.
Intergrating by parts, we get
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