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This mock test of Integral Calculus -3 for IIT JAM helps you for every IIT JAM entrance exam.
This contains 20 Multiple Choice Questions for IIT JAM Integral Calculus -3 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

The integral equals:

Solution:

We have

So,

implies

QUESTION: 2

is equal to:

Solution:

QUESTION: 3

Solution:

QUESTION: 4

is equal to:

Solution:

QUESTION: 5

is equal to:

Solution:

QUESTION: 6

The value of definite integral is equal to:

Solution:

QUESTION: 7

Let sin^{2} ucos^{2} udu, then:

Solution:

QUESTION: 8

is equal to:

Solution:

QUESTION: 9

is equal to:

Solution:

QUESTION: 10

Solution:

QUESTION: 11

The moment of inertia of a hollow sphere about a diameter is:

Solution:

Moment of Inertia of a Hollow Sphere about the Diameter

Suppose the mass of a hollow sphere is M, ρ is the density, inner radius R_{2} and outer radius R_{1},

∴ M = 4/3π(R_{1}^{3 }− R_{2}^{3})ρ

Moment of inertia of a hollow sphere (I) = Moment of inertia of a solid sphere of radius R_{1} - Moment of inertia of a solid sphere of radius R_{2}

QUESTION: 12

The area bounded by the curve from x = 0 to x = 1, the x -axis and the ordinate at x = 1 is:

Solution:

QUESTION: 13

The area of the closed curve x = a cos t, y = b sint is given by:

Solution:

x = acost , y = bsint

→ a^{2}/x^{2} + b^{2}/y^{2}=1

It's a hyperbola whose area in any quadrant is given by πab/4

QUESTION: 14

The area bounded by the curves y = e^{x}, y = e^{-x} and the line x = 1, is:

Solution:

Given curves are

y = e^{x}

y = e^{-}^{x }

From eq^{n} (i) and (ii), we get

QUESTION: 15

The intrinsic equation of the curve p = r sin α is given by:

Solution:

Given curve is p = r sin α ....(i)

Hence

or,

QUESTION: 16

The area of surface of the solid generated by the revolution of the line segment y = 2x from x = 0 to x = 2, about x-axis is equal to:

Solution:

Required surface area is

QUESTION: 17

The perimeter of the curve r = 2 cos θ is:

Solution:

The curve is r = 2 cos θ

Changing it into Cartesian coordinate, we get

x^{2} + y^{2} = 2x

or, (x - l)^{2} + y^{2} = 1,

which is a circle with centre (1,0) and radius 1.

Hence

Perimeter = 2πr = 2π · 1 = 2π

QUESTION: 18

The length of the arc of the parabola x^{2} = 4ay from the vertex to the extremity of the latus rectum is given by:

Solution:

Given curve is x^{2} = 4ay

Differentiating, w.r. to y, we get

implies

implies

Hence Length of the arc between vertex and extremity of latus rectum

QUESTION: 19

The area bounded by the curve y = x^{3}, x -axis and the line x = 1 and x = 4 is given by:

Solution:

Given curve is y = x^{3}

The required area = Area ABCD =

QUESTION: 20

The length of the arc of the curve y = log sec x between x = 0 and x = π/6 is equal to:

Solution:

Given curve is y = log sec x

So,

Hence Required length is given by

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