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

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## 45 Questions MCQ Test Mathematics For JEE | Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals

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Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 1
Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 1

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

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 2
Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 2

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 3
Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 3

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 4

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 4

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 5

If y = f(x) makes +ve intercept of 2 and 0 unit on x and y axes and encloses an area of 3/4 square unit with the axes then

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 5

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 6

The area bounded by the curves y = lnx, y = ln |x|,y=| ln x | and y = | ln |x| | is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 6

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.

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 7

The area of  the region bounded by the curves y = |x - 1| and y = 3 - |x| is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 7

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 8

is equal to

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 8

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 9

Let f(x)  be a function satisfying f '(x) = f(x) with f(0) = 1 and g(x) be a function that satisfies f(x)  + g(x) = x2 . Then the value of the integral

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 9

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 10

The value of the integral

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 10

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 11

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 11

[Using definite integrals as limit of sum]

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 12

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 12

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 13

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 13

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 14

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 14

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 15

and    then the value of   is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 15

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 16

The area of the region boun ded by the curves y = | x - 2 |, x = 1,x = 3 and the x-axis is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 16

The required area is shown by shaded region

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 17

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 17

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 18

The area enclosed between the curve y = loge (x +e) and the coordinate axes is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 18

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

e -e - 0 + 1=1

Hence the required area is 1 square unit.

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 19

The parabolas y2 = 4x and x2 = 4y divide the square region bounded by the lines x = 4, y = 4 and the coordinate axes. If S1 , S2 , S3 are respectively the areas of these parts numbered from top to bottom; then S1 : S2 : S3 is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 19

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

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 20

Let f (x) be a non – negative continuous function such that the area bounded by the curve y = f (x), x - axis and the
ordinates

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 20

Differentiating w. r . t β

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 21

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 21

Adding equations (1) and (2) we get

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 22

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 22

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 23

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 23

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 24

is equal to

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 24

[ using the property of even and odd function]

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 25

The value of  denotes the greatest integer not exceeding x is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 25

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])}

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 26

Then F(e) equals

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 26

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]

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 27

The solution for x of the equation

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 27

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 28

The area enclosed between the curves y2 = x and y = | x | is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 28

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

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 29

Then which one of the following is true?

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 29

for x∈(0, 1)

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 30

The area of the plane region bounded by the curves x + 2y2 = 0 and  x + 3y2 = 1is equal to

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 30

[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

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 31

The area of the region bounded by the parabola (y – 2)2 = x –1, the tangent of the parabola at the point (2, 3) and the x-axis is:

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 31

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)

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 32

denotes the greatest integer function, is equal to :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 32

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

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 33

The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 33

Area above x-axis = Area below x-axis

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 34

Let p(x) be a function defined on R such that p '(x) = p'(1 – x), for all x ∈ [0, 1], p (0) = 1 and p (1) = 41. Then

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 34

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

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

The value of

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 35

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 36

The area of the region enclosed by the curves y = x, x = e, y = 1/x and the positive x-axis is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 36

Area of required region AOBC

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 37

The area between the parabolas   and the straight line y = 2 is :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 37

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.

*Multiple options can be correct
Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 38

then  g (x + π) equals

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 38

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

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 39

Statement-1 : The value of the integral

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 39

Also, Given, I

…(2)

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

It is fundamental property.

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 40

The area (in square units) bounded by the curves  and lying in the first quadrant is :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 40

Given curves are

…(1)
and 2y – x + 3 = 0   …(2)

On solving both we get y = –1, 3

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 41

The integral

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 41

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 42

The area of the region described by

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 42

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)

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 43

The area (in sq. units) of the region described by {(x, y) : y2 < 2x and y > 4x – 1} is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 43

Required area

= Area of ABCD – ar (ABOCD)

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 44

The integral

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 44

`..(i)     ...(ii)

Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 45

The area (in sq. units) of the region {(x, y) : y2 > 2x and x2 + y2 < 4x, x > 0, y > 0} is :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Definite Integrals and Applications of Integrals - Question 45

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

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