Solving Calculus Problems and Linear Algebra Equations

This text provides solutions to calculus and linear algebra problems, including integration, distance and angle calculations, and verification of the Cauchy-Schwarz inequality. It also includes an example of verifying the parallelogram law and solving a system of linear equations.

3. (a)
⟨ f , g ⟩ =∫
a
b
f ( x ) g ( x ) dx
¿∫
−1
1
x2 e−x dx
integrating by parts we have ,letf =x2 , g'=e− x ⟹ f ' =2 x ,
g=−e−x .
∫
−1
1
x2 e− x dx =−x2 e− x−∫−2 x e−x dx .
solving for ∫−2 x e−x dx
−2∫ x e− x dx=−2 x e−x−2∫−e− x dx .
solving for ∫−e−x dx=e− x .
plugging∈the parts we have
∫
−1
1
x2 e− x dx =−x2 e− x−2 x e−x−2 e− x ¿−1
1
¿ e−5 e−1=0.8788
(b)
we define the length of a vector f by
‖f ‖= √ ⟨ f , f ⟩
which is simply thenorm .
This can be evaluated ¿
¿ √ ∫
−1
1
x2 x2 dx= √ ∫
−1
1
x4 dx
¿ √ 1
5 x5 ¿−1
1
¿ √ 1
5 − (−1
5 )
¿ 0.632 5
(c) the distance between f and g can be obtained from
d ( f , g )=‖f −g‖= √ ⟨ f −g , f −g ⟩ .
we need ¿ plug∈the f ∧g functions∧integrate . i. e
d ( f , g )=‖x2−e− x
‖= √ ⟨ x2−e−x , x2−e−x ⟩
¿ √ ∫
−1
1
( x2−e− x ) ( x2−e−x ) dx

¿ √ ∫
−1
1
(x ¿¿ 4−2 x2 e−x +e−2 x )dx ¿
¿ √ [2 x2 e−x +4 x e− x+ 4 e− x− e−x
2 + x5
5 ]−1
1
¿ 1.5064
(d) The angle between f and g can be calculated from the formula
cos θ= ⟨ f , g ⟩
‖f ‖‖g‖
now we need ¿ the|value of ‖g‖since ⟨ f , g ⟩ ∧‖f ‖have been calculated
¿ a∧b above .
‖g‖= √ ⟨ g , g ⟩ = √ ∫
−1
1
e− x e− x dx= √∫
−1
1
e−2 x dx .
we now let u=−2 x ⟹ dx=−1
2 du
‖g‖= √ −1
2 ∫ eu du= √−1
2 eu= √ [ −1
2 e−2 x
]−1
1
=1.90442
Now cos θ=¿ 0.87888
0.6325 ×1.90442 ⟹ θ=cos−1
( 0.87888
0.6325 ×1.90442 )=43.14 ° ¿
4.
¿ verify the Cauchy−s chwarz inequality , we need ¿ show that
f ( x ) =cos x∧g ( x ) =sin x where ⟨ f , g ⟩ =∫
0
π
4
cos x sin x dx
satisfies ⟨ f , g ⟩ 2 ≤ ⟨ f , f ⟩ ⟨ g , g ⟩ . Lets start by evaluating the LHS i . e
⟨ f , g ⟩ 2
= (∫
0
π
4
cos x sin x dx )
2
let u=sin x ⟹ dx= 1
cos x du ⟹∫u du= u2
2 =( [ sin2 x
2 ] 0
π
4
)
2
=0.0625
⟨ f , f ⟩= √∫
0
π
4
cos x cos x dx= √∫
0
π
4
cos2 x dx which can be solved using thereduction formula ¿ get
√∫
0
π
4
cos2 xdx= [ cos x sin x
2 + x
2 ]0
π
4 =0.8017

⟨ g , g ⟩ = √ ∫
0
π
4
sin x sin x dx= √∫
0
π
4
sin2 xdx which can be solved using thereduction formula ¿ get
√∫
0
π
4
sin2 xdx =[ −cos x sin x
2 + x
2 ]0
π
4 =0.3777 .
Now we have
0.0625<0.8017 × 0.3777=0.3028
thus satsfying theinequality ∎
5. To prove the parallelogram law
‖u+v‖2 +‖u−v‖2=2‖u‖2 +2 ‖v‖2
we can expand‖u+v‖2= ⟨ u+v ,u+ v ⟩ =‖u ‖2 + ⟨ u , v ⟩ + ⟨ u , v ⟩ +‖v‖2 =‖u‖2+ 2 ⟨ u , v ⟩ +‖v‖2 …… … …… . ( i )
‖u−v ‖
2= ⟨ u−v , u−v ⟩ =‖u ‖
2− ⟨ u , v ⟩ − ⟨ u , v ⟩ +‖v‖
2=‖u‖
2−2 ⟨ u , v ⟩ +‖v‖
2 … … … … … . ( ii )
Now adding ( i )∧ ( ii ) we have
‖u+v‖2 +‖u−v‖2=‖u‖2 +2 ⟨ u , v ⟩ +‖v‖2 +‖u‖2−2 ⟨ u , v ⟩ +‖v‖2 =2‖u‖2 +2 ‖v‖2
8. B= { ( 7,24,0,0 ) , ( 0,0,1,1 ) , ( 0,0,1 ,−2 ) }
Take u1= ( 7,24,0,0 ) ,u 2= ( 0,0,1,1 ) ,u 3= ( 0,0,1,−2 )
let v 1=u 1 , e 1= v 1
‖v 1‖ = ( 7,24,0,0 )
√ 72 +242+ 0+0 = ( 7
25 , 24
25 , 0,0 )
v 2=u 2− ⟨ u 2 , e 1 ⟩ e 1= ( 0,0,1,1 )− (0 × ( 7
25 , 24
25 , 0,0 ) )= ( 0,0,1,1 )
e 2= v 2
‖v 2‖= ( 0,0,1,1 )
√0+0+1+1 = (0,0 , 1
√2 , 1
√2 )
v 3=u 3− ⟨ u3 ,e 1 ⟩ e 1− ⟨ u 3 , e 2 ⟩ e 2= ( 0,0,1 ,−2 ) −0− ⟨ ( 0,0,1 ,−2 ) , (0,0 , 1
√2 , 1
√2 ) ⟩ (0,0 , 1
√2 , 1
√2 )=(0,0 ,
e 3= v 3
‖v 3‖= (0,0 , 3
2 , −3
2 )
√0+ 0+( 3
2 )2
+ (−3
2 )2 =(0,0 , √2
2 ,− √ 2
2 )

B' = {( 7
25 , 24
25 , 0,0 ) , ( 0,0 , 1
√ 2 , 1
√ 2 ) , ( 0,0 , √ 2
2 ,− √ 2
2 ) }
9. X1
( 3
5
4
5
0 ) + X2
( −4
5
3
5
0 ) + X2 ( 0
0
1 )= ( 10
25
5 ) we can find the values of X1 , X 2 , X3
3
5 X1 − 4
5 X2+ 0 X3=10 … … … . ( i )
4
5 X1 + 3
5 X2 +0 X 3=25 … … ..(ii)
X3 =5 …… …(iii)
¿ ( i )∧ ( ii ) we have
3 X1−4 X2 =50 , 4 X1 +3 X2 =125 , solving X1=26 , X2=7 hence XB =
[ 26
7
5 ]
References:
Seymour (1976) .Schaum’s outline Linear Algebra 4th ed. Mac-GrawHill
Gilbert s.(1993) Introduction to linear algebra 2nd ed.
Sheldon A.(1995). Linear Algebra done right. Springer
Steven R.(1992).Advanced Linear algebra. Springer.