Examples
(1) Given
d
y
d
x
=
24
x
11
{\displaystyle {\dfrac {dy}{dx}}=24x^{11}}
. Find
y
{\displaystyle y}
. Ans .
y
=
x
3
2
+
C
{\displaystyle y=x^{3}2+C}
.
(2) Find
∫
(
a
+
b
)
(
x
+
1
)
d
x
{\displaystyle \int (a+b)(x+1)\,dx}
. It is
(
a
+
b
)
∫
(
x
+
1
)
d
x
{\displaystyle (a+b)\int (x+1)\,dx}
or
(
a
+
b
)
[
∫
x
d
x
+
∫
d
x
]
{\displaystyle (a+b)\left[\int x\,dx+\int dx\right]}
or
(
a
+
b
)
(
x
2
2
+
x
)
+
C
{\displaystyle (a+b)\left({\dfrac {x^{2}}{2}}+x\right)+C}
.
(3) Given
d
u
d
t
=
g
t
1
2
{\displaystyle {\dfrac {du}{dt}}=gt^{\frac {1}{2}}}
. Find
u
{\displaystyle u}
. Ans .
u
=
2
3
g
t
3
2
+
C
{\displaystyle u={\frac {2}{3}}gt^{\frac {3}{2}}+C}
.
(4)
d
y
d
x
=
x
3
−
x
2
+
x
{\displaystyle {\dfrac {dy}{dx}}=x^{3}-x^{2}+x}
. Find
y
{\displaystyle y}
.
d
y
=
(
x
3
−
x
2
+
x
)
d
x
or
d
y
=
x
3
d
x
−
x
2
d
x
+
x
d
x
;
y
=
∫
x
3
d
x
−
∫
x
2
d
x
+
∫
x
d
x
;
{\displaystyle {\begin{aligned}dy&=(x^{3}-x^{2}+x)\,dx\quad {\text{or}}\\dy&=x^{3}\,dx-x^{2}\,dx+x\,dx;\quad y=\int x^{3}\,dx-\int x^{2}\,dx+\int x\,dx;\end{aligned}}}
and
y
=
1
4
x
4
−
1
3
x
3
+
1
2
x
2
+
C
{\displaystyle \quad \quad \quad \quad y={\tfrac {1}{4}}x^{4}-{\tfrac {1}{3}}x^{3}+{\tfrac {1}{2}}x^{2}+C}
.
(5) Integrate
9.75
x
2.25
d
x
{\displaystyle 9.75x^{2.25}\,dx}
. Ans .
y
=
3
x
3.25
+
C
{\displaystyle y=3x^{3.25}+C}
.
All these are easy enough. Let us try another case.
Let
d
y
d
x
=
a
x
−
1
{\displaystyle \quad \quad \quad \quad {\dfrac {dy}{dx}}=ax^{-1}}
.
Proceeding as before, we will write
d
y
=
a
x
−
1
⋅
d
x
,
∫
d
y
=
a
∫
x
−
1
d
x
{\displaystyle dy=ax^{-1}\cdot dx,\quad \int dy=a\int x^{-1}\,dx}
.
Well, but what is the integral of
x
−
1
d
x
{\displaystyle x^{-1}\,dx}
?
If you look back amongst the results of differentiating
x
2
{\displaystyle x^{2}}
and
x
3
{\displaystyle x^{3}}
and
x
n
{\displaystyle x^{n}}
, etc., you will find we never got
x
−
1
{\displaystyle x^{-1}}
from any one of them as the value of
d
y
d
x
{\displaystyle {\dfrac {dy}{dx}}}
. We got
3
x
2
{\displaystyle 3x^{2}}
from
x
3
{\displaystyle x^{3}}
; we got
2
x
{\displaystyle 2x}
from
x
2
{\displaystyle x^{2}}
; we got
1
{\displaystyle 1}
from
x
1
{\displaystyle x^{1}}
(that is, from
x
{\displaystyle x}
itself); but we did not get
x
−
1
{\displaystyle x^{-1}}
from
x
0
{\displaystyle x^{0}}
, for two very good reasons. First ,
x
0
{\displaystyle x^{0}}
is simply =1, and is a constant, and could not have