Self learning activity

1. CIA SELF LEARNING ACTIVITY SUB-
MATHEMATICS- II
SWAMI GROUP NO - 6
DURING THEACADEMIC YEAR 2023-24

2. Q1.Take the photo of increasing and decreasing curves separately. Verify whether
Roll's theorem is applicable or not, if applicable take approximate value and find C

3. f(x)=𝑥2 + 2x − 8 ;x𝜖[-4,2]
lets check conditions of rolls theorem
condition 1 and 2-
f(x)=𝑥2+2x-8 is a polynomial so f(x)is continuous on [-4,2] and
differentiable in(-4,2)
Condition 3-
f(a)=f(b)
for a=-4,b=2
f(-4)=(−4)2+2 −4 − 8
=16-8-8
=0
f(2)=(2)2+2 2 − 8
=4+4-8
=0

4. Hence, f(-4) = f(2)
Now
f(x) = 𝑥2 + 2x – 8 f ' (x)
= 2x + 2 – 0 f ' (x) = 2x
+2
put x=c
f ' (c) = 2c +2
Since all three conditions are satisfied
f ' (c) = o
2c + 2 = 0
2c = -2
c = -2/2 c=-1
hence the rolls theorem is verified.

5. Q2. Solve 𝑑𝑦
= 𝑋 + 𝑋2𝑌2, 𝑑𝑧
= 𝑋𝑍, 𝑌 1 = 3, 𝑍 1 = 4, ℎ = 0.3
𝑑𝑥 𝑑𝑥
Find 𝑦, 𝑧 at 𝑥 = 1.3
𝑌0 = 3, 𝑋0 = 1, 𝑍0 = 4, ℎ = 0.3
Here,
𝑑𝑦
= 𝑓 𝑥, 𝑦, 𝑧 = 𝑥 + 𝑥2𝑦2 & 𝑑𝑧
= 𝜑 𝑥, 𝑦, 𝑧 = 𝑥𝑧
𝑑𝑥 𝑑𝑥
Starting at (𝑥0, 𝑦0, 𝑧0) and taking the step-sizes for 𝑥,𝑦, 𝑧 to be ℎ, 𝑘, 𝑙 respectively,
the Runge-Kutta method gives
𝑘1 = ℎ𝑓(𝑥0,𝑦0, 𝑧0)
0 0
= ℎ (𝑥0 + 𝑥2𝑦2)
= 0.3 (1 + 1232)
= 0.3 10
𝑙1 = ℎ𝜑(𝑥0, 𝑦0, 𝑧0)
= ℎ (𝑥0𝑧0)
= 0.3(1 × 4)
= 1.2
𝑘1 = 3

6. 𝑘2 2
0 0
= ℎ𝑓 𝑥 +
ℎ
,𝑦 +
𝑘1
2 0
, 𝑧 +
𝑙
1
2
ℎ 𝑘1
𝑙2 = ℎ𝜑(𝑥0 + 2
, 𝑦0 + 2
,𝑧0 2
+
𝑙1
)
= ℎ𝑓 ( 1 +
0.3
, 3 +
2 2
1.2
2
3
, 4 + )
3
2 2
= 0.3𝜑(1 +
0.3
,3 + , 4 +
2
1.2
)
= ℎ𝑓( 1.15,4.5,4.6 ) = 0.3𝜑 (1.15,4.5,4.6)
= 0.3 ( 1.15 + 1.15 2 4.5 2) = 0.3 (1.15 × 4.6)
= 0.3 (1.15 + 1.3225× 20.25) = 0.3 (5.29)
= 0.3 (27.9306) = 1.587
= 8.3792

7. 𝑘3 2
0 0
= ℎ𝑓 𝑥 + ℎ
,𝑦 +
𝑘1
2 0
, 𝑧 +
𝑙2
2 3
ℎ
0 0
𝑙 = ℎ𝜑(𝑥 + , 𝑦 +
𝑘2
0
2 2 2
, 𝑧 + 𝑙2
)
= 0.3𝑓 ( 1.15,3 +
2
8.3792
, 4 +
2
1.587
) = 0.3𝜑 (1.15,7.1896,4.7935)
= 0.3𝑓( 1.15,14.3792
,
2 2
9.857
) = 0.3 (1.15 × 4.7935)
= 0.3𝑓 ( 1.15,7.1896,4.7935) = 0.3 (5.5125)
= 0.3 (1.15 + (1.15)2(7.1896)2) = 1.65385
= 0.3 (69.5104)
= 20.85312

8. 𝑙4 = ℎ𝜑(𝑥0 + ℎ, 𝑦0 + 𝑘3, 𝑧0 + 𝑙3 )
𝑘4 = ℎ𝑓 𝑥0 + ℎ, 𝑦0 + 𝑘3, 𝑧0 + 𝑙3
= 0.3𝑓 ( 1 + 0.3,3 + 20.85312,4 + 1.6538) = 0.3𝜑 (1.3,23.85312,5.6538)
= 0.3𝑓( 1.3,23.85312,5.6538) = 0.3 (1.3 × 5.6538)
= 0.3 (1.3 + (1.3)2(23.85312)2) = 2.2050
= 0.3 (962.8614)
= 288.85842

9. 1
𝑙 = (𝑙1 + 2𝑙2 + 2𝑙3 + 𝑙4 )
6
1
= ( 3 + 2 8.3792 + 2 20.85312 + 288.8584)
1
6
= (1.2 + 2 1.537
6
2 1.6538 + 2.2050)
1
= (3 + 41.70624 + 16.7584 + 288.8584) =
1
6
(1.2 + 3.174 +
6
3.3076 + 2.2050)
1
6
= (350.32306) =
1
6
(9.8866)
= 58.3871 = 1.6477
𝑦 = 𝑦0 + 𝑘 = 3 + 58.3571
= 5.6477
𝑧 = 𝑧0 + 𝑙 = 4 + 1.6477
= 61.38
K =1 (K1 + 2K2 + 2K3 + K4 )
6
+ 2 (1.65385)+ 2.2050 )
3.3077+ 2.2050 )

10. Q.3. Determine Area lies under the C shape curve, take photo of such curve you have
seen in college campus.

11. eq-1
eq-2
Parabola : 𝑦2 = 4𝑥
Line : 𝑦 = 2𝑥 − 4
Put the equation 1 in equation 2
∴ (2𝑥 − 4)2= 4𝑥
∴ 4𝑥2 − 16𝑥 + 16 = 4𝑥
∴ 4𝑥2 − 20𝑥 + 16 = 0
∴ 𝑥2 −5𝑥 + 4 = 0
𝑥 = 1,4
When x = 1 and
When x = 4 and
y = -2
y = 4
Point of intersection are A(1,-2) and B(4,4)
2
Internal limit: 𝑥 = 𝑦
4
; 𝑥 =
(𝑦+4)
2
External limits : y = -2 ; y = 4

12. A = 𝑦=−2
𝑦=4
𝑥=
𝑦2
4
𝑥=
(𝑦+4)
2
1 𝑑𝑥 𝑑𝑦
𝐴 =
−2
4
𝑥 𝑦2
4
(𝑦+4)
2
𝑑𝑦
𝐴 =
−2
4
𝑦 + 4
2
−
𝑦2
4
𝑑𝑦
𝐴 =
−2
4
2𝑦 + 8 − 𝑦2
4
𝑑𝑦
𝐴 =
1
4 −2
4
(2𝑦 + 8 − 𝑦2
) 𝑑𝑦
𝐴 =
1
4
2𝑦2
2
+ 8𝑦 − 𝑦3/3
−2
4

13. 𝐴 42 + 8 4
23
)
A 16 + 32
(( 2)2 + 8 2
8
3
(4 16 + )
A
3
64
3
4 + 16
8
3
A
A
16 + 32
60 24
36
A = 9 sq. units

14. Q.4. Using tylor’s series express (𝒙 − 𝟓)𝟑+2(𝒙 − 𝟓)𝟐 +
7 in power of‘x’.
Ans:
f(x)= 𝒙 − 𝟓 𝟑+2(𝒙 − 𝟓)𝟐+7
f(x)=𝒙𝟑 +𝟐𝒙𝟐+7
h= -5
f(x+h)= f(h)+xf’(h)+𝒙𝟐/2!.f”(h)+𝒙𝟑
/3!+f”(h) f(-5)=(−𝟓)𝟑+2X(−𝟓)𝟐+7
= (-125)+50+7
.: f(-5)= -68
F`(h)= 𝟑𝒙𝟐+4x
F`(-5)=3(−𝟓)𝟐+4(-5)
=55

15. F``(h)=6x+4
f``(-5)=6(-5)+4
= -30+4
.: f’”(h)= -26
f’”(h)= 6
f’”(-5)= 6
f(x+h)= -68+x(55)+𝒙𝟐/2.(-26)+𝒙𝟑/6 x(6)
= -68+55x-13𝒙𝟐+𝒙𝟑
= 𝒙𝟑-13𝒙𝟐+55x-68
.: f(x+h)= 𝒙𝟑-13x^2+55x-68

16. THANK YOU!!!