UNIT 2

Question-1: Evaluate

.Sol. Here we notice that f:x→cos x is a decreasing function on [a , b],

Therefore by the definition of the definite integrals-

Then

Now,

Here

Thus

Question-2: Evaluate 0∞ x3/2 e -x dx

Solution: 0∞ x3/2 e -x dx = 0∞ x 5/2-1 e -x dx

= γ (5/2)

= γ (3/2+ 1)

= 3/2 γ (3/2 )

= 3/2. ½ γ (½ )

= 3/2 .½.π

= ¾ π

Question-3: Determine the area enclosed by the curves-

Sol. We know that the curves are equal at the points of interaction, thus equating the values of y of each curve-

Which gives-

By factorization,

Which means,

x = 2 and x = -3

By determining the intersection points the range the values of x has been found-

x | -3 | -2 | -1 | 0 | 1 | 2 |

1 | 10 | 5 | 2 | 1 | 2 | 5 |

And

x | -3 | 0 | 2 |

y = 7 - x | 10 | 7 | 5 |

We get the following figure by using above two tables-

Area of shaded region =

=

= ( 12 – 2 – 8/3 ) – (-18 – 9/2 + 9)

=

= 125/6 square unit

Question-4: Evaluate I =

Solution:

= 2 π/3

Question-5: Find the area enclosed by the two functions-

and g(x) = 6 – x

Sol. We get the following figure by using these two equations

To find the intersection points of two functions f(x) and g(x)-

f(x) = g(x)

On factorizing, we get-

x = 6, -2

Now

Then, area under the curve-

A =

Therefore the area under the curve is 64/3 square unit.

Question-6: Find the volume of the solid of revolution formed by revolving R around y-axis of the function f(x) = 1/x over the interval [1 , 3].

Sol. The graph of the function f(x) = 1/x will look like-

The volume of the solid of revolution generated by revolving R(violet region) about the y-axis over the interval [1 , 3]

Then the volume of the solid will be-

Question-7: Find the volume of the solid of revolution formed by revolving R around y-axis of the function f(x) = 2x - x² over the interval [0 , 2].

Sol. The graph of the function f(x) = 2x - x² will be-

The volume of the solid is given by-

=

Question-8: Evaluate , where dA is the small area in xy-plane.

Sol.

Let, I =

=

=

=

= 84 sq. Unit.

Which is the required area.

Question-9: Evaluate

Sol. Let us suppose the integral is I,

I =

Put c = 1 – x in I, we get

I =

Suppose , y = ct

then dy = c

Now we get,

I =

I =

I =

I =

I =

As we know that by beta function,

Which gives,

Now put the value of c, we get

Question-10: Evaluate ey/x dy dx.

Sol.

Given: I = ey/x dy dx

Here limits of inner integral are functions of y therefore integrate w.r.t y,

I = dx

=

=

I =

= =

ey/x dy dx=

Question-11: Evaluate

Sol.

Let,I =

Here limits for both x and y are constants, the integral can be evaluated first w.r.t any of the variables x or y.

I = dy

I=

=

=

=

=

=

=

Question-12: Evaluate over x 1, y

Sol.

Let I= over x 1, y

The region bounded by x 1 and y

Is as shown in Fig. 6.3.

Fig. 6.3

Take a vertical strip along strip x constant and y varies from y =

To y = . Now slide strip throughout region keeping parallel to y-axis. Therefore y constant and x varies from x = 1 to x = .

I=

=

= [ ∵ dx = tan–1 (x/a)]

= =

=– = (0 – 1)

I =

Question-13: Evaluate

Soln. :

The region of integration bounded by

y = 0, y = and x = 0, x = 1

y = x2 + y2 = x

The region bounded by these is as shown in Fig

Convert the integration in polar co-ordinates by using x = r cos , y = r sin and dx dy = r dr d

x2 + y2= x becomes r = cos

y = 0 becomes r sin = 0 = 0

x = 0 becomes r cos = 0 =

And x = 1 becomes r = sec Fig. 6.8

Take a radial strip SR with angular thickness , Along strip constant and r varies from r = 0 to r = cos . Turning strip throughout region therefore varies from = 0 to =

I = r dr d

= 4 cos sin d r dr

= 4 cos sin d [–]

= – 2 cos sin [+1] d

= – 2 [cos sin – cos sin ] d

= –2 + 2 cos sin d

= – + 2

= + 1 =

Question-14: Evaluate r sin dr d over the cardioid r = a (1 – cos ) above the initial line.

Soln. :

The cardioid r = a (1 – cos ) is as shown in Fig. 6.11. The region of the integration is above the initial line.

Take a radial strip SR, along strip constant and r Varies from r = 0 to r = a (1 – cos ).

New turning the strip throughout region therefore varies from = 0 to = .

I= r sin dr d

= sin d

Fig.6.11

=sin [a2 (1 – cos )2]

=

I= (sin – 2 sin cos + sin ) d

= 2 (sin – sin2 + sin ) d

=+

I= a2= a2

I=

Question-15: Find the area enclosed by two curves using double integration.

y = 2 – x and y² = 2 (2 – x)

Sol. Let,

y = 2 – x ………………..(1)

And y² = 2 (2 – x) ………………..(2)

On solving eq. (1) and (2)

We get the intersection points (2,0) and (0,2) ,

We know that,

Area =

Here we will find the area as below,

Area =

Which gives,

= ( - 4 + 4 /2 ) + 8 / 3 = 2 / 3.

Question-16: Find the are lying inside a cardioid r = 1 + cos θ and ouside the parabola r(1 + cos θ) = 1.

Sol. Let,

r = 1 + cos θ ……………………..(1)

r(1 + cos θ) = 1……………………..(2)

Solving these equestions , we get

(1 + cos θ )( 1 + cos θ ) = 1

(1 + cos θ )² = 1

1 + cos θ = 1

Cos θ = 0

θ = ±π / 2

So that, limits of r are,

1 + cos θ and 1 / 1 + cos θ

The area can be founded as below,

Question-17: Evaluate

Solution:-

Question-18: If the density at any point of a non-uniform circular lamina of radius’ a’ varies as its distance from a fixed point on the circumference of the circle then find the mass of lamina.

Solution:

Take the fixed point on the circumference of the circle as origin and diameter through it as x axis. The polar equation of circle

And density.