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# Subjective Type Questions: Definite Integrals and Applications of Integrals - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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Q. 1. Find the area bounded by the curve x2 = 4y and the straight line x = 4y – 2.           (1981 - 4 Marks)

Ans.

Solution. To find the area bounded by

x2 = 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

x2 = x+2 ⇒ x2 -x -2 = 0 ⇒ (x-2)(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 x2 – y2 = 1. Show that the area bounded by this hyperbola and the lines joining its centre to  the points corresponding to t1 and –t1 is t1.          (1982 - 3 Marks)

Solution. Let P(t1) and Q(–t1) 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 x-axis, 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 x-axis 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, x2 + y2 = 25, 4y = | 4 – x2 | and x = 0 above the x-axis.          (1987 - 6 Marks)

Ans.

Solution. We have to find the area bounded by the curves

x2 +y2 = 25               ...(1)

...(2)

x = 0  ...(3)
and above x-axis.

Thus we have three curves

or (y - 2)2 = 52 ∴ y - 2 = ±5 y = 7,y =-3
y = – 3, 7 are rejected since.
y = – 3 is below x-axis and
y = 7 gives imaginary value of x. So, (I) and (II) do not intersect but II intersects x-axis at (2, 0) and (–2, 0). (I) and (III) intersect at

y = – 7 is rejected, y = 3 gives the points above x-axis. 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 x-axis.        (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. co-ordinates 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 x-axis 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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