Integration as Reverse Differentiation [ edit | edit source ] The power rule for differentiation states that:
d
d
x
x
n
=
n
x
n
−
1
{\displaystyle {\dfrac {d}{dx}}x^{n}=nx^{n-1}}
The reverse of this is
∫
n
x
n
−
1
d
x
=
x
n
+
c
{\displaystyle \int nx^{n-1}dx=x^{n}+c}
∫
x
n
d
x
=
x
n
+
1
n
+
1
+
c
{\displaystyle \int x^{n}dx={\frac {x^{n+1}}{n+1}}+c}
This is the power rule for integration.
Regarding the derivatives of exponentials and logarithms, we know that:
d
d
x
e
x
=
e
x
{\displaystyle {\dfrac {d}{dx}}e^{x}=e^{x}}
d
d
x
ln
x
=
1
x
{\displaystyle {\dfrac {d}{dx}}\ln x={\frac {1}{x}}}
Reversing these, we get:
∫
e
x
=
e
x
+
c
{\displaystyle \int e^{x}=e^{x}+c}
∫
1
x
=
ln
|
x
|
{\displaystyle \int {\frac {1}{x}}=\ln |x|}
The natural logarithm takes a modulus input so that it can handle negative numbers.
Note that similar rules apply to any linear expressions that may be composed with these functions:
∫
e
a
x
+
b
=
1
a
e
a
x
+
b
+
c
{\displaystyle \int e^{ax+b}={\frac {1}{a}}e^{ax+b}+c}
∫
1
a
x
+
b
=
1
a
ln
|
a
x
+
b
|
{\displaystyle \int {\frac {1}{ax+b}}={\frac {1}{a}}\ln |ax+b|}
The derivatives of the trigonometric functions are:
d
d
x
sin
x
=
cos
x
{\displaystyle {\dfrac {d}{dx}}\sin x=\cos x}
d
d
x
cos
x
=
−
sin
x
{\displaystyle {\dfrac {d}{dx}}\cos x=-\sin x}
d
d
x
tan
x
=
sec
2
x
{\displaystyle {\dfrac {d}{dx}}\tan x=\sec ^{2}x}
d
d
x
tan
−
1
x
=
1
1
+
x
2
{\displaystyle {\dfrac {d}{dx}}\tan ^{-1}x={\frac {1}{1+x^{2}}}}
Thus, the corresponding integrals are:
∫
cos
x
=
sin
x
+
c
{\displaystyle \int \cos x=\sin x+c}
∫
sin
x
=
−
cos
x
+
c
{\displaystyle \int \sin x=-\cos x+c}
∫
sec
2
x
=
tan
x
+
c
{\displaystyle \int \sec ^{2}x=\tan x+c}
∫
1
1
+
x
2
=
tan
−
1
x
+
c
{\displaystyle \int {\frac {1}{1+x^{2}}}=\tan ^{-1}x+c}
As with exponentials and logarithms, this applies to linear expressions that are composed with trigonometric functions:
∫
cos
(
a
x
+
b
)
=
1
a
sin
(
a
x
+
b
)
+
c
{\displaystyle \int \cos(ax+b)={\frac {1}{a}}\sin(ax+b)+c}
∫
sin
(
a
x
+
b
)
=
−
1
a
cos
(
a
x
+
b
)
+
c
{\displaystyle \int \sin(ax+b)=-{\frac {1}{a}}\cos(ax+b)+c}
∫
sec
2
(
a
x
+
b
)
=
1
a
tan
(
a
x
+
b
)
+
c
{\displaystyle \int \sec ^{2}(ax+b)={\frac {1}{a}}\tan(ax+b)+c}
∫
1
1
+
(
a
x
+
b
)
2
=
1
a
tan
−
1
(
a
x
+
b
)
+
c
{\displaystyle \int {\frac {1}{1+(ax+b)^{2}}}={\frac {1}{a}}\tan ^{-1}(ax+b)+c}
Using Trigonometry when Integrating [ edit | edit source ] Sometimes, it is useful to use trigonometric identities when finding the integral of an expression.
e.g. Find the integral
∫
4
cos
2
x
d
x
{\displaystyle \int 4\cos ^{2}xdx}
cos
2
x
=
2
cos
2
x
−
1
Double Angle Identity
cos
2
x
=
cos
2
x
−
1
2
∫
4
cos
2
x
d
x
=
∫
4
cos
2
x
−
1
2
d
x
=
∫
cos
2
x
−
1
d
x
=
1
2
sin
2
x
−
x
+
c
{\displaystyle {\begin{aligned}\cos 2x&=2\cos ^{2}x-1\quad {\text{Double Angle Identity}}\\\cos ^{2}x&={\frac {\cos 2x-1}{2}}\\\int 4\cos ^{2}xdx&=\int 4{\frac {\cos 2x-1}{2}}dx\\&=\int \cos 2x-1dx\\&={\frac {1}{2}}\sin 2x-x+c\end{aligned}}}
The area under a curve can be approximated using the area of several trapeziums The trapezium rule states that the area under a curve can be approximated by finding the sum of the areas of trapeziums. The area of a trapezium is given by
A
=
B
+
b
2
h
{\displaystyle A={\frac {B+b}{2}}h}
B
{\displaystyle B}
b
{\displaystyle b}
h
{\displaystyle h}
In the context of the trapezium rule, each trapezium's perpendicular distance
h
{\displaystyle h}
Δ
x
{\displaystyle \Delta x}
f
(
x
)
{\displaystyle f(x)}
A
i
=
f
(
x
i
)
+
f
(
x
i
+
1
)
2
Δ
x
{\displaystyle A_{i}={\frac {f(x_{i})+f(x_{i+1})}{2}}\Delta x}
The area under the curve between the bounds
a
{\displaystyle a}
b
{\displaystyle b}
A
=
∑
x
=
a
b
f
(
x
)
+
f
(
x
+
Δ
x
)
2
Δ
x
{\displaystyle A=\sum _{x=a}^{b}{\frac {f(x)+f(x+\Delta x)}{2}}\Delta x}
Because
Δ
x
{\displaystyle \Delta x}
A
=
f
(
a
)
+
2
f
(
a
+
Δ
x
)
+
2
f
(
a
+
2
Δ
x
)
+
⋯
+
2
f
(
b
−
Δ
x
)
+
(
b
)
2
Δ
x
{\displaystyle A={\frac {f(a)+2f(a+\Delta x)+2f(a+2\Delta x)+\dots +2f(b-\Delta x)+(b)}{2}}\Delta x}
A
=
(
f
(
a
)
+
f
(
b
)
2
+
f
(
a
+
Δ
x
)
+
f
(
a
+
2
Δ
x
)
+
⋯
+
f
(
b
−
Δ
x
)
)
Δ
x
{\displaystyle A=\left({\frac {f(a)+f(b)}{2}}+f(a+\Delta x)+f(a+2\Delta x)+\dots +f(b-\Delta x)\right)\Delta x}
Thus, the trapezium rule states:
∫
a
b
f
(
x
)
d
x
≈
(
f
(
a
)
+
f
(
b
)
2
+
f
(
a
+
Δ
x
)
+
f
(
a
+
2
Δ
x
)
+
⋯
+
f
(
b
−
Δ
x
)
)
Δ
x
{\displaystyle \int _{a}^{b}f(x)dx\approx \left({\frac {f(a)+f(b)}{2}}+f(a+\Delta x)+f(a+2\Delta x)+\dots +f(b-\Delta x)\right)\Delta x}
Trapezium Rule Worked Example [ edit | edit source ] Q. Use the trapezium rule with 4 intervals to estimate the value of
∫
0
4
2
4
x
−
x
2
d
x
{\textstyle \int _{0}^{4}2{\sqrt {4x-x^{2}}}~dx}
A.
a
=
0
{\textstyle a=0}
b
=
4
{\textstyle b=4}
h
=
1
{\textstyle h=1}
Using
y
=
2
4
x
−
x
2
{\displaystyle y=2{\sqrt {4x-x^{2}}}}
x
{\textstyle x}
0
1
2
3
4
y
{\textstyle y}
y
0
=
0
{\textstyle y_{0}=0}
y
1
=
2
3
{\textstyle y_{1}=2{\sqrt {3}}}
y
2
=
4
{\textstyle y_{2}={4}}
y
3
=
2
3
{\textstyle y_{3}=2{\sqrt {3}}}
y
4
=
0
{\textstyle y_{4}=0}
∫
0
4
2
4
x
−
x
2
d
x
≈
h
2
[
y
0
+
y
4
+
2
(
y
1
+
y
2
+
y
3
)
]
{\displaystyle \int _{0}^{4}2{\sqrt {4x-x^{2}}}~dx\approx {\frac {h}{2}}\left[y_{0}+y_{4}+2\left(y_{1}+y_{2}+y_{3}\right)\right]}
≈
1
2
[
0
+
0
+
2
(
2
3
+
4
+
2
3
)
]
{\displaystyle \approx {\frac {1}{2}}[0+0+2(2{\sqrt {3}}+4+2{\sqrt {3}})]}
≈
4
3
+
4
{\displaystyle \approx 4{\sqrt {3}}+4}
≈
10.93
{\displaystyle \approx 10.93}
← Differentiation · Numerical Solution of Equations →
![{\displaystyle \int _{0}^{4}2{\sqrt {4x-x^{2}}}~dx\approx {\frac {h}{2}}\left[y_{0}+y_{4}+2\left(y_{1}+y_{2}+y_{3}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38c7ace428f0b0cbdc28aa5f16d0bfc1eda2aef5)
![{\displaystyle \approx {\frac {1}{2}}[0+0+2(2{\sqrt {3}}+4+2{\sqrt {3}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4745e3a0c07d4f238717712226d003600a20d569)
