The document contains examples of indefinite integrals of various functions: 1) Finding the antiderivatives of polynomials like 4x3 + 3x - 2. 2) Finding an antiderivative involving a differential equation like dy/dx = 4x3 - 4x. 3) Evaluating integrals involving rational functions like ∫(3 - 2/x2 + 6x3)dx. 4) Finding antiderivatives of expressions involving radicals like ∫((x2+3)2/x2)dx. 5) Solving differential equations and evaluating integrals using substitution.

1. Findthe indefinite integral foreachof the
following
a) ∫(4𝑥3 + 3𝑥 − 2 )dx
=
4𝑥4
4
+
3𝑥2
2
– 2x + C
=𝑥4 +
3𝑥2
2
− 2𝑥 + 𝐶
b)
𝑑𝑦
𝑑𝑥
= 4𝑥3 − 4𝑥, y=0 whenx=2 .
Findy intermsof x
∫
𝑑𝑦
𝑑𝑥
𝑑𝑥 = ∫4𝑥3 − 4𝑥 dx
Y =
4𝑥3+1
4
-
4𝑥1+1
2
+ C
= 𝑥4 - 2𝑥2+ C
0= 24 - 2(22)+ C
0 = 16 -8 + C
-8= C
y = 𝑥4 - 2𝑥2 - 8
c) ∫(3 −
2
𝑥2
+
6
𝑥3
)dx
∫(3 − 2𝑥−2 + 6𝑥−3) dx
= 3x –
2𝑥−2+1
−1
+
6𝑥−3+1
−2
+ C
= 3x +
2𝑥−1
1
-
3𝑥−2
1
+ C
= 3x + 2𝑥−1 - 3𝑥−2 + C
d) ∫(
(𝑥2+3)2
𝑥2
)dx
= (
(𝑥2 + 3)
𝑥2
(𝑥2 + 3)
𝑑𝑥
= (
𝑥4 + 3𝑥2 + 3𝑥2+9
𝑥2
) dx
= (𝑥2 + 6 + 9 𝑥−2) dx
=
𝑥3
3
+ 6x + –
9𝑥−1
−1
+ C
=
𝑥3
3
+ 6x - –
9𝑥−1
1
+ C
1 𝑑𝑝
𝑑𝑣
= 2v -
𝑣3
2
,p=0 whenv=0
findthe value of p whenv=1
2
Giventhat
𝑑2
𝑦
𝑑𝑥2 = 4x and that
𝑑𝑦
𝑑𝑥
= 0 ,
Y=2 whenx=0 . Find
𝑑𝑦
𝑑𝑥
and y interms
𝑑𝑝
𝑑𝑣
= 2v -
𝑣3
2 𝑑2
𝑦
𝑑𝑥2 = 4x
dp = ( 2v -
𝑣3
2
) dv ∫ 𝑑
2
𝑦 = ∫4x 𝑑𝑥2
= 2vdv -
𝑣3
2
dy=
4𝑥2
2
dx
∫ dp = ∫2vdv −
𝑣3
2
dv
𝑑𝑦
𝑑𝑥
=
4𝑥2
2
= 2𝑥2
p = 2
𝑣2
2
-
𝑣3+1
2(4)
+ C
= 𝑣2 -
𝑣4
8
+ C
0 = 02 -
04
8
+ C p= 12 -
14
8
+ C
c = 0 = 1 – 1/8 + 0
= 7/8
2𝑥2 =0
X= 0
∫
𝑑𝑦
𝑑𝑥
= ∫ 2𝑥2
y =
2𝑥3
3
+ C =
2(0)3
3
+ C
2 = C
y =
2𝑥3
3
+ 2

2. 3 Findthe equationof the curve withgradiet
2𝑥2 + 3𝑥 − 1
𝑑𝑦
𝑑𝑥
= 2𝑥2 + 3𝑥 − 1
∫ 𝑑𝑦 = ∫2𝑥2 + 3𝑥 − 1 dx
y = 2
𝑥3
3
+ 3
𝑥2
3
- x + C
4
Giventhat
∫ 𝑓 ( 𝑥) 𝑑𝑥 = 3 𝑎𝑛𝑑
2
1
∫ 𝑓 ( 𝑥) 𝑑𝑥 = −7
2
3
a) The value of k if ∫ [ 𝑘𝑥 − 𝑓 ( 𝑥)] 𝑑𝑥 = 8
2
1
b) ∫ [5𝑓( 𝑥) − 1] 𝑑𝑥
2
1
a) ∫ [ 𝑘𝑥 − 𝑓 ( 𝑥)] 𝑑𝑥 = 8
2
1
∫ [ 𝑘𝑥𝑑𝑥 ] −
2
1
∫ [ 𝑓( 𝑥) 𝑑𝑥 ] = 8
2
1
k [
𝑥2
2
] − 3= 8
k(
22
2
−
12
2
)=11
k ( 2-
1
2
) = 11
k (
3
2
) = 11
3k= 22
K=
22
3
b) ∫ [5𝑓( 𝑥) − 1] 𝑑𝑥
3
1
= 5 [∫ 𝑑𝑥
2
1 + ∫ 𝑑𝑥 ]
3
2 - [ x]1
3
= 5 [ 3 + 7 ]– ( 3-1)
= 50 -2= 48
a) ∫5 (2 − 3𝑣)−6dv
=
5(2−3𝑣)−6+1
−6+1(−3)
+ C
=
5(2−3𝑣)−5
−5 (−3)
+ C
=
(2−3𝑣)−5
− (−3)
+ C
=
1(2−3𝑣)−5
3
+ C
b)∫
4
3(1−5𝑥)5
dx
=
4
3
∫(1 − 5𝑥)−5dx
=
4
3
[
(1−5𝑥)−5+1
(−5+1)(−5)
] + C
=
4
3
[
(1−5𝑥)−4
(−4)(−5)
] + C
=
4
3
[
(1−5𝑥)−4
20
] + C
=
1
15
[ (1 − 5𝑥)−4] + C