Test: JEE Main 35 Year PYQs- Definite Integrals & Applications of Integrals - Airforce X Y / Indian Navy SSR MCQ

<[∵ | 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