Solution of Assignment ( Order No: 933942)

Solution of Assignment ( Order No: 933942)
1. Solve 2 cosh 2 x+10 sinh2 x=5
Solution:
We use definition of hyperbolic functions:
cosh 2 x= e2 x +e−2 x
2 , sinh 2 x= e2 x−e−2 x
2 . Therefore given equation
becomes
2 ( e2 x +e−2 x
2 )+10 ( e2 x−e−2 x
2 )=5 .... After cancelling
e2 x +e−2 x +5 e2 x−5 e−2 x=5
6 e2 x−4 e−2 x=5
6 e2 x− 4
e2 x =5 ..... Now take LCM
6 e2 x e2 x−4=5 e2 x
6 e4 x−5 e2 x−4=0. Now we put e2 x=t
6 t2−5 t−4=0 ... This is quadratic equation in t .
t=−(−5) ± √ 25+ 4 × 6× 4
2 ×6 ...Simplifying we get
t= 4
3 and t=−1
2 ... we backsubstitute value of t
∴ e2 x= 4
3 or e2 x=−1
2 which is not possible as exponential function is
never negative.
∴ e2 x= 4
3 that implies 2 x=log ( 4
3 )
∴ x =1
2 log ( 4
3 )
2. Let f be differentiable on ( a , b ) and let c ∈ ( a ,b ). Show that
( a−b ) f ( c ) +∫
a
b
f ( x ) dx=∫
a
b
∫
c
x
f ' ( y ) dydx
Solution:
R.H.S.= ∫
a
b
∫
c
x
f ' ( y ) dydx
= ∫
a
b
[ f ( x ) −f (c) ] dx .....From first fundamental theorem of calculus
= ∫
a
b
f ( x ) dx−∫
a
b
f ( c ) dx by separating integral

= ∫
a
b
f ( x ) dx−f (c)∫
a
b
dx
= ∫
a
b
f ( x ) dx−f (c)(b−a)
= ∫
a
b
f ( x ) dx + f (c )(a−b)
= L.H.S.
3. Evaluate the double integral ∫∫
D
y
x2 + y2 dA
Where D is the region in the lower half plane lying between the circles x2+ y2=1
And x2+ y2=2
Solution: We use here polar coordinates;
x=rcosθ , y=rsinθ , x2+ y2=r2 and dxdy =rdrdθ
Therefore given integral becomes
I = ∫
θ=0
π
∫
r =1
r = √ 2
rsinθ
r2 rdrdθ ........ cancelling r and integrating with respect to r ,we get
I = ∫
θ=0
π
[ r ]1
√ 2 sinθdθ ......... putting upper and lower limits
I = ( √ 2−1 ) ∫
0
π
sinθ dθ
I =¿ ( √2−1 ) [ −cosθ ]0
π .... Putting limits
I = ( √ 2−1 ) [ −cos ( π ) +cos (0) ]
I =2( √ 2−1)
4. Evaluate the iterated integral
∫
0
π
∫
y
π
sinx
x dxdy
Solution : We use here Fundamental theorem of Calculus
There exist function F ( y) such that F ( y ) =−∫
π
y
sinx
x dx and F' ( y )= siny
y
Let I =∫
0
π
∫
y
π
sinx
x dxdy

I =−∫
0
π
∫
π
y
sinx
x dxdy
I =−∫
0
π
1. F ( y ) dy ........ After putting value of F(y)
Integrating by parts
I =− [F ( y )∫
0
π
1 dy−∫
0
π
[ F' ( y ) .∫ 1 dy ] dy ]
I =− [ F ( y ) . y ] 0
π
+∫
0
π
[ siny
y . y ] dy
I =− [ F ( π ) . π −0 ] + [ −cos ( y ) ] 0
π
I =− [ 0−0 ] − [ cos ( π ) −cos ( 0 ) ] =− [ −1−1 ]
I =2
5. Given that z=f ( x , y ) is a plane in R3 and
∫∫
D
f ( x , y ) dA=∫∫
D
xf ( x , y ) dA=0 and f ( 1,2 ) =−1
Since f (x , y ) is a plane in R3. Let f ( x , y ) =ax +by +d is equation of plane.
Given that ∫∫
D
f ( x , y ) dA=0
∫
x=0
x=3
∫
y=0
y=3 −x
( ax +by +d ) dydx=0 ..... integrating with respect to y
∫
x=0
3
[axy + b y2
2 +dy ]0
3− y
dx =0 .... Putting limits
∫
0
3
[3 ax−a x2+ b ( 9−6 x + x2 )
2 +3 d −xd ]dx=0 .....integrating w.r.t .x
[ 3 a x2
2 − a x3
3 + 9bx
2 − 3 x2 b
2 + b x3
6 +3 dx− d x2
2 ]0
3
= 0 ... putting limits