Calculus Mathematics Solution 2022

Running head: CALCULUS 1
Calculus
Student Name
Institution

CALCULUS 2
Question 1
Step 1: Identify the critical points by equating the derivative to zero
F(x) = x3+x2-5x+2
dy
dx = 3 x2 +2x-5
3 x2 +2x-5 = 0
Use the equation:
X=−b ± √b2−4 ac
2a = −2± √ 22−4∗3∗−5
2∗3 =−5
3 , 1
Step 2: find the domain
Since this is a quadratic equation, the domain is: −∞<¿x¿ ∞
Step 3: check the behavior of the monotone intervals
Since the critical points fall within the domain, we form the monotone intervals as:
−∞<¿X← 5
3 , −5
3 <x<1, 1<x< ∞
For −∞<¿x← 5
3 ;
We pick a value within the range (x=-1) and check the sign value of dy
dx
F(x) = 3(−1)2 +2(-1)-5=-4
For −5
3 <x<1;
We pick a value within the range (x=0) and check the sign value of dy
dx
F(x) = 3(0)2 +2(0)-5=-5
We pick a value within the range (x=2) and check the sign value of dy
dx

CALCULUS 3
F(x) = 3(2)2 +2(2)-5=11
−∞ <¿x← 5
3
−5
3 <x<1 1<x<∞
Sign - - +
Behavior decreasing Decreasing Increasing
Question 2
Step 1: Find the first derivative
f(x) = x−2
x+ 1
Using quotient rule of differentiation;
d
dx ( x −2 ) ( x +1 ) − d
dx ( x +1 ) ( x −2 )
( x +1)2
= 1 ( x +1 ) −1( x−2)
(x +1)2 = 3
( x+1 )2
f ‘(x) = 3
( x+1 )2
Step 1: equate first derivative to zero and find where the first derivative is undefined
3
( x+1 ) 2 =0
The function 3
( x+1 )2 is undefined at x=-1 since we cannot divide by zero
Step 2: finding the domain
Domain: x<-1 or x>-1
Step 3: finding singularity points
x¿-1
Since f(x) is not defined at x¿-1, then no inflection points. Inflection points exists where a
function, f(x) is defined at all points within the domain (Hayter, 2013; Stephens, 2011).

CALCULUS 4
Question 3
Question 4
Step 1: Perimeter of the fence:
L+W = 80
Step 2: Making L the subject of the formula
L = 80-w
Step 3: Calculate the area
Area = L*W
= (80-W)W

CALCULUS 5
A = 80W-W 2
Step 4: differentiate and equate to zero
dA
dW = 80-2W
80-2W=0
W= 40
L= 80-W = 80-40 = 40
Maximum area = L*W = 40*40 = 1600 ft2
Question 5
Step 1: expressing width in terms of x
32 = 2x+w
W= 32-2x
Step 2: calculating area with width (w) expressed in x
Area, A = L*W
A = x*w
A = x(32-2x) = 32x - 2x2
Step 3: calculating the x values which maximizes cross-sectional area by differentiating area (A)
function
dA
dx = 32 – 4x
Step 4: equating dA
dx to zero
32 – 4x = 0
4x= 32
Thus

CALCULUS 6
X= 8
W= 32-2x = 32-2(8) = 16
Step 4: checking whether the extreme point is maximum by calculating the second derivative of
area function
d2 A
d x2 = – 4
Thus the extreme point is maximum
Step 5: calculating the maximum cross-sectional area
A = L*W
= 16*8
= 128 inch2
Question 6
a)
( 6 x8 ) ( 3 x−4 ) = 18( x−4 ) ( x8 )
Using the power rule, we simplify
18( x−4 ) ( x8 )= 18( x−4 +8 )=18 ( x4 )
Therefore,
( 6 x8 ) ( 3 x−4 ) = 18( x4 )
b)
3 a−5
27 a−7 = a−5
9 a−7
= 1
9 a(−5− (−7 ))
= 1
9 a(−5 +7)

CALCULUS 7
= 1
9 a2
= ( 1
3 a )
2
Question 7
log x
3 x +1
Step 1: apply logarithm rule: logc ( a
b ) = logc ( a ) -logc ( b )
log ( x
3 x +1 ) = log10 ( x )-log10 ( 3 x+1 )
=log10 ( x )-log10 ( 3 x+1 )
Question 8
a)
Q(t) = 410e0.0035t
At t=0;
Q(t) = 410 e0.0035(0)= 410
b)
At t=50;
Q(t) = 410 e0.0035(50) = 488.411
Question 9

CALCULUS 8
y=ex x2lnx
Let f= ex , g=x2 ln (x)
Apply product rule: (f ∙ g)’ = f' ∙g +f∙g
d
dx (ex ) x2 lnx + d
dx ( x2 lnx ) ex ………………………………………………… Equation (1)
But
d
dx (ex ) = ex
d
dx (ex ) x2 lnx= ¿) x2 lnx …………………………………………………...Equation (2)
For d
dx ( x2 lnx ) ex ,
Apply exponent rule: ab = ebln(a)
x2 ln (x) = e2 ln ( x ) ln (x)
d
dx ( x2 lnx ) ex = d
dx (e2 ln ( x ) ln (x) )
Then, apply chain rule to differentiate ( x2 lnx ) ex
Let x= eu and d
dx ( 2 ln ( x ) ln ( x) )
d
du (eu ¿ = eu
And
d
dx ( 2 ln ( x ) ln (x) )= 2( 1
x ln ( x ) + 1
x ln ( x ) )
Substitute u = 2ln(x)ln(x)
e2 ln ( x ) ln (x) * 2( 1
x ln ( x )+ 1
x ln ( x ) )e x ……. …………………………………Equation (3)

CALCULUS 9
Adding equation (2) and (3) gives;
¿) x2 lnx+e2 ln ( x ) ln (x) * 2( 1
x ln ( x )+ 1
x ln ( x ) )e x
Therefore,
d
dx (ex x2 lnx )= ¿) x2 lnx+e2 ln ( x ) ln (x) * 2( 1
x ln ( x )+ 1
x ln ( x ) )e x
Question 10
∫ ( 25 x4 +4 x +3 )dx = 25∫ x4dx +4∫ xdx +3∫ 1dx
= 25[ x4 +1
4+1 ]+ 4[ x1+1
1+1 ] + 3 [ x0 +1
0+1 ]+C
=25 [ x5
5 ]+ 4 [ x2
2 ] + 3[ x
1 ]+C
= 5 x5+2x2+3x +C
Question 11
P=-0.01 x2 – 0.2x+10
Calculating the Equilibrium price when x=0
dP
dx = -0.02x-0.2
x=-10
Demand when x=-10
P=-0.01(−10)2 – 0.2(-10)+10 = $11
Demand when x=2
P=-0.01(2)2 – 0.2(2)+10 = $9.56
Consumer surplus = 11-9.56 = $1.44
Question 12

CALCULUS 10
∫t (t +2)−6dt
Step 1 : Simplify (t +2)−6
(t +2)−6 = t
(t +2)6
Thus
∫t (t +2)−6dt = ∫ t
( t+2)6 dt
Step 2: use substitution method of integration
Let u=t+2
t=u-2
Substitute t+2 with u
∫ t
( t+2)6 dt = ∫ u−2
(u)6 du where u=t+2 and t=u-2
Step 3: expand u−2
(u)6
u−2
(u)6 = 1
u5 - 2
u6
∫ ( 1
u5 − 2
u6 ) du = ∫ 1
u5 du - ∫ 2
u6 du
= ∫u−5du - 2∫ u−6du
= −1
4 u4 - ( −2
5 u2 )
Step 3: substitute u=t+2 in −1
4 u4 - ( −2
5 u2 )
−1
4 u4 - ( −2
5 u2 ) = −1
4 (t+ 2)4 +( 2
5(t +2)2 ) +C

CALCULUS 11
Therefore,
∫t (t +2)−6dt = −1
4 (t+ 2)4 + ( 2
5(t +2)2 ) +C
References
Hayter, A. (2013). Probability and statistics for engineers and scientists. Australia: Duxbury.
Stephens, L. (2011). Statistics for engineers. New York: McGraw-Hill Companies.

CALCULUS 12