Calculus Solved Problems with Examples - Desklib

This page contains solved problems in calculus with examples covering topics such as derivatives, integrals, optimization, and more. The content includes solutions to problems in differentiation, integration, optimization, and related topics. The subject is Calculus, and the course code is not mentioned. The course name and college/university are not mentioned.

CALCULUS
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Question 1
a.
7 x2+ 2
x3 +5 3
√ x−1
dy
dx ()=7 d
dx ( x2 )+2 d
dx (x ¿¿−3)+5 d
dx (x
1
3 )−0 ¿
¿ 7 ( 2 ) ( x2−1 ) +2(−3)( x¿¿−3−1)+5 ( 1
3 ) x
1
3 −1
¿
¿ 14 x−6 x− 4+ 5
3 x
−2
3
b. y=In ( 3 x−2 )
Let 3 x−2=u , so that y=lnu
dy
du = 1
u , du
dx =3
dy
dx = dy
du × du
dx = 1
u × 3= 3
u = 3
3 x−2
c. y=5 sin 3 x
dy
dx = d
dx ¿
Let 3 x=u , so that y=sinu
dy
du =cosu , du
dx =3
dy
dx =5 dy
du × du
dx =5 ( cosu ) (3 )=15 cosu=15 cos (3 x)
d.
y= 3 e2 x
x3
We apply the quotient rule to obtain
dy
dx =
x3 d
dx ( 3 e2 x ) −3 e2 x d
dx (x3 )
(x3 )2
¿ x3 (6 e¿¿ 2 x)−3 e2 x(3 x2)
x6 ¿
¿ 6 x3 e2 x−9 x2 e2 x
x6 = 6 x e2 x−9 e2 x
x4
e. y=x3 cos 4 x
We apply the product rule to obtain
dy
dx =x3 d
dx ( cos 4 x )+ cos 4 x d
dx ( x ¿¿ 3)¿
¿ x3 ( −4 sin 4 x ) + cos 4 x (3 x ¿¿ 2) ¿
¿−4 x3 sin (4 x )+3 x2 cos (4 x )
¿ 3 x2 cos ( 4 x)−4 x3 sin ( 4 x)
f.
y= ( x2−7 ) 6
Let x2−7=u , so that du
dx =2 x ,∧ y=u6
dy
du = d
du (u¿¿ 6)=6u5 ¿
dy
dx = dy
du × du
dx =6u5 × 2 x=12 x u5=12 x ( x2−7)5
Question 2
Part a
y=x3 e2 x
at x = 2
We apply the product rule as follows
dy
dx =x3 d
dx (e2 x )+e2 x d
dx ( x3 )
¿ x3 ( 2 e2 x ) +e2 x ( 3 x2 )
dy
dx =2 x3 e2 x+3 x2 e2 x

When x=2
dy
dx =2(2)3 e2(2)+3 ( 2 )2 e2 (2 )=16 e4 +12 e4=e4 ( 16+12 )=28 e4=1528.75
Part b
y=sin 3 x
x2 at x= π
4
We apply the quotient rule to obtain
dy
dx =
x2 d
dx ( sin ( 3 x ) ) −sin ( 3 x ) d
dx (x2)
(x2)2
¿ x2 ( 3 cos ( 3 x ) )−sin ( 3 x ) (2 x )
x4
¿ 3 x cos ( 3 x ) −2 sin(3 x)
x3
When x= π
4
dy
dx =
3 π
4 cos (3 π
4 )−2sin ( 3 π
4 )
( π
4 )
3
¿
3 π
4 ( − √ 2
2 ) −2 ( √ 2
2 )
π
64
3 =
−64 ( 3 π √ 2
8 − √ 2)
π3
¿ −8 √ 2 ( 3 π +8 )
π3 =−6.358
Question 3
y=2 x3−9 x2+ 12 x +1
To get the turning points, we determine the derivative of the
curve and then equate it to zero as follows:
dy
dx =0= d
dx ( 2 x3−9 x2 +12 x +1 )
d
dx ( 2 x3 −9 x2+12 x+ 1 )=6 x2−18 x+ 12=0
a=6 , b=−18 , c=12
x= 18 ± √ (−18)2 −4 (6 ×12)
2× 6 =18 ± √ 324−288
12 = 18 ± 6
12 = 24
12
x= 24
12 =2∨x=12
12 =1
x=1∨x =2
when x=1 , y=2(1)3−9 ( 1 )2+ 12 ( 1 ) +1=6
when x=2 , y=2(2)3−9 ( 2 ) 2 +12 ( 2 ) +1=5
Hence, the turning points are ( 1 , 6 )∧(2 ,5)
To determine whether the points are maximum, we choose
values on either sides of the turning points and evaluate the
derivatives as summarized in table 1 below.
dy
dx =6 x2−18 x+12
When x=0 x=1when x=1.5 x=2 when x=3
dy
dx
12( positive)0 −1.5( negative)0 12( positive)
Sig
n ¿ −¿ ¿ −¿ ¿
Therefore, we can see that ( 1 , 6 )is a maximum point whereas
(2 , 5) is a minimum point.
Question 4
x=2 t3−15 t2+24 t +4
Part a
The velocity , v= dx
dt = d
dt (2 t3 −15 t2 +24 t +4)
¿ 6 t2−30 t+24
When t=5 , v =6(5)2−30 ( 5 ) +24=24 m/ s
Part b

acceleration= dv
dt = d
dt ( 6 t2−30 t +24)
¿ 12 t −30
When t=1 , acceleration=12 ( 1 ) −30=−18 m/s2
Part c
At rest , velocity=0. That is,
6 t2−30 t+24=0 , a=6 , b=−30 , c=24
t= 30 ± √(−30)2−4 (6 ×24)
2 ×12 = 30 ± √900−576
24 = 30 ±18
24 = 48
24 ∨12
24
∴ t=2 sec∨t=0.5 sec
Part d
From part b, acceleration=12t −30
When acceleration=6 m/s2 we can get the time as follows
12 t−30=6
12 t=6+30=36
12t
12 =36
12 =3 ,t=3 sec
Question 5
E=1.16−(5.232× 10−3 )T −(7 ×10−6) T2
To get the maximum E, we equate the derivative of the
function to zero as follows
dE
dt =0=0−5.232× 10−3−2 ¿
2 ¿
T =−5.232× 10−3
2 ¿¿
Maximum
E=1.16−(5.232× 10−3 )(−373.7143)−(7× 10−6 )(−373.7143)2
Emax=1.16+1.95527−0.97764=2.13763
Question 8
volume of cylinder=π r2 h=1000, h=1000
π r2
S . A of cylinder=2 π r 2+2 πrh=2 π r 2+2 πr ( 1000
π r2 )
S . A=2 π r 2+ 2000
r
To minimize area, d ( S . A )=0. That is,
d
dr ( 2 π r2 + 2000
r )=0
4 πr −2000
r2 =0
( 4 πr− 2000
r2 ) r2=0 (r2)
4 π r3 −2000=0
4 π r3 =2000
r3 =2000
4 π = 500
π
radius=3
√ 500
π =5.42 cm
We get the corresponding height as
h=1000
π r2 =h= 1000
π (5.42)2 =10.84 cm
Question 9
a. ∫5 e2 x + 3
√ x4 dx
∫5 e2 x + 3
√ x4
dx=5∫e2 x dx +∫ x
4
3 dx
¿ 5 ( 1
2 e
2 x
)+ x
4
3 +1
4
3 +1
+C
¿ 5
2 e2 x + x
7
3
7
3
+C

¿ 5
2 e2 x + 3
7 x
7
3 + C
b. ∫ 3
4 x dx
∫ 3
4 x dx= 3
4 ∫ 1
x dx= 3
4 ln ( x )+C
c. ∫ 4 √ x dx
∫ 4 √ x dx=4∫ x
1
2 dx=4 × x
1
2 +1
1
2 +1
+C
¿ 4 × x
3
2
3
2
+C=8
3 x
3
2 +C
d. ∫
0
π
cos x+sin 2 x dx
∫
0
π
cos x+sin 2 x dx=∫
0
π
cos xdx +∫
0
π
sin 2 x dx
¿|sin ( x)|0
π
+|−1
2 cos ( 2 x)|0
π
¿ ( 0−0 )− 1
2 ( cos ( 2 π ) −cos ( 0 ) )
¿−1
2 ( 1−1 )=0
e. ∫2 x ln x dx
∫2 x ln x dx=2∫ x ln x dx
We apply the Chain rule :∫udv=uv−∫ vdu as follows
We let x=u so that du=dx, and let
lnx=dv so that v =∫lnxdx=1/ x
uv−∫ vdu=x ( 1
x )−∫ ( 1
x )dx
¿ 1−ln ( x ) +C
Question 10
a. ∫
0
1
( 5 x−6 ) 5 dx
Let 5 x−6=u , dx= 1
5 du
∫
0
1
( 5 x−6 )5 dx =∫
0
1
1
5 u
5
du
∫ 1
5 u
5
= 1
5 ( 1
6 u
6
)= 1
30 u
6
= 1
30 (5 x−6)6
∫
0
1
( 5 x−6 )5 dx =| 1
30 ( 5 x−6 )6
|0
1
¿ 1
30 ((5(1)−6)¿¿ 6−( 5(0)−6)6 ) ¿
¿ 1
30 ((−1)¿¿ 6−(−6)6 )¿
¿ 1
30 ( 1−46656 )=−1555.167
b. ∫
0
π
4
x sinx dx
We apply the Chain rule :∫udv=uv−∫ vdu as follows
We let x=u so that du=dx, and let
sinx=dv so that v=∫ sinxdx=−cosx
uv−∫ vdu=x ( −cosx ) −∫ ( −cosx ) dx
¿−xcosx+sinx
uv−∫ vdu=sinx−xcosx
∫
0
π
4
x sinx dx=|sinx−xcosx|0
π
4
¿ ¿
¿ 1
√2 (1− π
4 )=0.15175
c. ∫
0
1
12−4 x
x2+2 x−3 dx
First, we evaluate the function into partial fraction as follows
12−4 x
x2 +2 x−3 = 12−4 x
x2+3 x−x−3 = 12−4 x
x (+3)−1( x +3)= 12−4 x
( x−1)(x+ 3)
12−4 x
( x−1)(x +3)= A
(x−1) + B
(x +3)

12−4 x= A ( x +3 ) + B(x−1)
When x=1
12−4(1)= A ( 1+3 ) + B( 1−1)
8=4 A , A=2
When x=−3
12−4(−3)= A (−3+3 ) + B(−3−1)
24=−4 B , B=−6
∴ 12−4 x
(x−1)( x+3)= A
(x−1) + B
(x+3)= 2
( x−1)+ −6
( x+3)
Then we find the integral and evaluate the limits as follows
∫
0
1
12−4 x
x2+2 x−3 dx=∫
0
1
2
(x−1) dx +∫
0
1
−6
( x +3) dx
¿ 2∫
0
1
1
( x−1) dx−6∫
0
1
1
(x +3) dx
¿ 2 |ln ( x−1)|0
1
−6|ln (x +3)|0
1
¿ 2 { ln 0−ln ( −1 ) }−6 {ln 4−ln 3 }
¿ 2 {ln ( 0
−1 ) }−6 {ln ( 4
3 ) }
¿ ln ( 0 )=∞
Question 11
From the above diagram, the cone formed has the form x= y2
Therefore, the volume is given by V =π ∫
x=0
5
y2 dx
But we know that y=2 x so that
V =π ∫
x=0
5
(2 x)2 dx=4 π ∫
x=0
5
x2 dx
V =4 π |x3
3 |0
5
=4 π ( 53
3 )=4 π ( 125
3 )=166.667 π =523.6
Question 12
y=100 sin (2 θ) , 0 ≤ 2θ ≤ π
y RMS= √ 1
π−0 ∫
0
π
y2 dθ
¿ √ 1
π ∫
0
π
104 sin 2 (2 θ)dθ
¿ √ 104
π ∫
0
π
sin 2(2θ) dθ
We can write sin2 ( 2θ ) dθ= 1−cos ( 4 θ)
2 so that

√ 104
π ∫
0
π
sin 2 (2θ)dθ= √ 104
π ∫
0
π
1−cos (4 θ)
2 dθ
¿ √ 104
2 π ∫
0
π
1−cos (4 θ)dθ
¿ √ 104
2 π |θ− 1
4 sin (4 θ)|0
π
¿ √ 104
2 π ¿ ¿
RMS voltage= √ 104
2 π ( π )= √ 104
2 =70.7106V
Question 13
The region bounded by the given function and the specified
lines is shown in the figure above. To get the area, we
integrate the function y=x2−x for 1 ≤ x ≤ 2 then subtract
the area of trapezium formed below y=x2 from x=1 ¿ x=2
as follows.
Area=∫
1
2
x2 dx− ( 1
2 ( 2+1 ) × 1 )
¿| x3
3 |1
2
− 3
2 = 8
3 − 1
3 − 3
2 =5
6
Question 14
dθ
dt =kθ
dθ
kθ =dt , 1
kθ dθ=dt
Integrating both sides of the equation we obtain,
∫ 1
kθ dθ=∫ dt
1
k ∫ 1
θ dθ=∫dt
1
k ln ( θ )=t+C
ln ( θ )=k (t +C)
When t=2.5 min , t=65 ℃
ln ( 65 ) =k (2.5+ C) … equation1
When t=5 min , t=50℃
ln ( 50 ) =k (5+C )… equation 2
Then, we divide equation 1 by equation 2 to get
ln ( 65 )
ln ( 50 ) = k ( 2.5+C )
k ( 5+C )
1.067066=2.5+C
5+C
1.067066(5+C )=2.5+C
1.067066 C−C=2.5−5.33533
0.067066 C=−2.83533
C=−42.2767
Substituting C=−42.2767 into equation 1 we get
ln ( 65 ) =k (2.5−42.2767)
k = ln ( 65 )
(2.5−42.2767)= ln ( 65 )
−39.7767 =−0.1049
Therefore, ln ( θ )=k (t +C) becomes
ln ( θ ) =−0.1049(t−42.2767)
When θ=35℃ ,ln ( 35 ) =−0.1049(t−42.2767)

ln ( 35 )
−0.1049 =(t−42.2767)
−33.878044=t−42.2767
t=42.2767−33.878044=8.398656 minutes
t= ( 8.398656 ×60 ) s=503.91936
t ≅ 504 seconds