Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals

<[∵ | sin x| is periodic with period π and sin x > 0 if 0 < x π]

First we draw each curve as separate graph

NOTE : Graph of y = | f(x) | can be obtained from the graph of the curve y = f(x) by drawing the mirror image of the portion of the graph below x-axis, with respect to x-axis.
Clearly the bounded area is as shown in the following figure.

 [Using definite integrals as limit of sum]

The required area is shown by shaded region

The graph of the curve y = log_e (x+e) is as  shown in the fig.
 

e -e - 0 + 1=1

Hence the required area is 1 square unit.

Intersection points of x² = 4y and y² = 4x are (0, 0) and (4, 4). The graph is as shown in the figure.

∴ S₁ : S₂ :S₃ = 1 : 1:1

Differentiating w. r . t β

 



Adding equations (1) and (2) we get

 
Adding equation (1) and (2)

[ using the property of even and odd function]

Let a = k + h where k is an integer such that [a] =k and 0 < h < 1

{f (2) – f (1)} + 2{f (3) – f (2)} + 3{f  (4) –f (3)} + ........ + (k – 1) {f(k) – f(k – 1)} + k{f(k + h)  – f(k)}
= – f (1) – f (2) – f (3) ......... – f (k) + kf (k + h)
= [a] f (a) - {f (1) + f (2) + f (3)+ .......... f ([a])}

and limit for t = 1 ⇒ z = 1 and for t = 1/e ⇒ z = e

    [∴ log1 = 0]

Equation (A) becomes

Let log t = x 
[for limit t = 1, x = 0 and t = e, x = log e = 1]

The area enclosed between the curves y² = x and y = | x |
From the figure, area lies between y² = x and y = x

for x∈(0, 1)

[Left handed  parabola with vertex at (0, 0)]

[Left handed parabola with vertex at (1, 0)] Solving the two equations we get the points of intersection as (–2, 1), (–2, –1)

The required area is ACBDA, given by

The given parabola is (y – 2)² = x – 1 Vertex (1, 2) and it meets x–axis at (5, 0) Also it gives y² – 4y – x + 5 = 0
So, that equation of tangent to the parabola at (2, 3) is

which meets x-axis at (– 4, 0).
In the figure shaded area is the required area.
Let us draw PD perpendicular to y – axis.

Then required area = Ar ΔBOA + Ar (OCPD) – Ar (ΔAPD)
 

Adding two values of I in eqⁿ s (1) & (2), We get

Area above x-axis = Area below x-axis

p '( x) = p '(1-x) ⇒ p( x) = - p(1 - x)+c
at x = 0
p(0) = – p(1) + c ⇒ 42 = c
Now, p(x) = - p(1 -x)+ 42 ⇒ p (x) + p (1 -x) = 42

on adding (i) and (ii), 

Area of required region AOBC

Given curves     are the parabolas whose equations can be written as  y = 4x² and

Also, given y = 2. 

Now, shaded portion shows the required area which is symmetric.

Putting t = π + y in second integral, we get

Also, Given, I

            …(2)

By adding (1) and (2), we get

It is fundamental property.

Given curves are

             …(1)
and 2y – x + 3 = 0   …(2)

On solving both we get y = –1, 3

Given curves are x² + y² = 1 and y² = 1 –x. Intersecting points are x = 0, 1

Area of shaded portion is the required area.
So, Required Area =  Area of semi-circle + Area bounded by parabola

(∵ radius of circle = 1)

Required area

= Area of ABCD – ar (ABOCD)

     `..(i)     ...(ii)
Adding (1) and (2)

Points of intersection of the two curves are (0, 0), (2, 2) and (2, –2)
Area = Area (OAB) – area under parabola (0 to 2)