Monash University MAT1841 Assignment 3: Continuous Mathematics for CS

This assignment solution addresses problems from MAT1841, focusing on continuous mathematics for computer science. The solution includes calculations for the first three non-zero terms of a series, determining coefficients for an equation based on given points, solving an integral using integration by parts, sketching a function, and applying the fundamental theorem of calculus. Furthermore, the solution provides calculations using the trapezoidal rule for numerical integration. The assignment demonstrates the application of calculus principles and numerical methods to solve mathematical problems.

Question 1
a. f ( x ) =ex3
, a=1
The series is given as follows;
f ( x ) =Σn =0
∞ ¿
= f ( a ) +f ' ( a ) ( x−a)+ f ″ ( a )
2! ( x−a)2 + f ‴ ( a )
3 ! ( x−a)3 + …
Where f ( a ) , f ' (a ) ( x−a ) , f ″ ( a )
2 ! (x−a)2 , f ‴ ( a )
3 ! ( x−a)3 +…
forms the first, the second and the third term respective for the
series
Now obtaining the first value on the series
f ( x )=exp ( x3 ) so f(a) = 2.72
Obtaining the second value
f ' ( x )=3 x2 e x3
so f ' ( 1 )=3 e x2
the third value
f ″ ( x )=6 x ex3
(1+ 3
2 x3 ) so f ″ ( 1 )=15 ex3
Now plugging the above into the first equation
= 2.72+3 ex2
(x−a)+ 15 ex3
2 ! (x−a)2 these are 1st,2nd and 3rd values of series
b. f ( x )=tan−1 ( e−x−1),a=0

The first constant is given as follows
f ( x )=tan−1 ( e−x−1) so f ( 0 )=0
The second value is given as follows
f ' ( x )= −e−x
( e− x−1 )2
+ 1 so f ' ( 0 )=1
f ″ ( x ) = ex ( 2 e2 x−1 )
( 2 e2 x−2 ex+1 ) 2 so f ″ ( 0 )=1
Now plugging this values into the first equation
= 0 + x + 1
2! x2
Question 2
(1, -2), (2, -1), (4, -0.5) and (8, -0.25)
They take the following form
p1 ( x ) =σ1 x3 +γ1 x2 +β1 x +α1
p2 ( x ) =σ2 x3 + γ2 x2+β2 x+ α2
the first condition require
p1 ( 1 ) = -2

p1 ( 2 ) = -1
p2 ( 1 ) = -0.5
p2 ( 2 ) = -0.25
the second condition
p1
' (2) = p2
' (2)
the condition requires
p1 ″ (2) = p2 ″ (2)
the last condition
p1 ″ (1) =0
p1 ″ (3) =0
there eight equations with 8 unknowns therefore

Upon solving the following is the solution
So the equation now is
p1 ( x ) =1.25 x3−11.5 x2+ 31.50 x −19
Question 3
a. I =∫ x2 ln ( x ) dx
∫uv′=uv−∫u′v

u=ln ( x ) v =x2
u'= 1
x , v = x3
3
nowbecomes
I = x3 ln ( x )
3 −∫ x2
3 dx
= x3 ln ( x )
3 - x3
9 +C
b. I= ∫ e3 x sin ( x ) dx
∫ uv ' = uv− ∫ u' v
u=sin x, v' =e3 x
u'=cos ( x ) , v=e3 x
3
I = e3 x sin x
3 −∫ ( e3 x cos ( x )
3 ) dx
The second part has to be integrated by part as well
u=cos x , u'=−sin x

v'= e3 x
3 , v= e3 x
9 so
I = e3 x sin x
3 −( e3 x cos x
9 −∫ (−e3 x sin ( x )
9 )dx )
Now become
I = e3 x ( sin ( x )−cos ( x ) )
10 +C
Question 4
a. The sketch
The sketch of y=2x2−8 x and y= x3 −5 x2 +4 x
5
b. The fundamental theorem

∫
a
b
f ( x ) −g(x) dx = F( x ) ∨a
b = F(b) -F(a)
so for equation 1
I= ∫
0
4
x3−5 x2+ 4 x
5 −2 x2 +8 x dx
= ∫ 0.2 x3−3 x2+ 8.8 x dx
= 0.2 x4 −x3 + 4.4 x2∨4
0
= 0.2 x 44−43 +4.4(42 )
= 51.2 – 64+ 70.4
= 57.6
c. Trapezoidal rule for n=2 using 0.2 x3−3 x2 +8.8 x
Area = b−a
2 n [ f ( a ) +2 f ( x1 ) +2 f ( x2 ) +…+ f ( b ) ]
A=0, b=4 therefore
Area = b−a
2 n [ f ( 0 ) +2 f ( 2 ) +f ( 4 ) ]

= 4−0
2∗2 ( 0 + 7.2 +0)
= 7.2
d. Area for n= 4
= 4−0
2∗4 ( 0 +6+ 7.2 +0)
= 6.6