My Goal
My goal is to make mathematics more accessible and fun for everyone, and a big part of that is to explain mathematics using "easy language", but this requires a balancing act between precision and comprehension.
Let me explain: there is an educational concept called the spiral, which roughly means that a subject comes around again and again, always at a higher level. For example, a young person is taught that multiplication is just repeated addition. But then a year later the subject is revisited and multiplying by negatives is taught, then decimals come along ...
File:Multiply-p2n3.gif This is an illustration of 2 times -3. Observe that our toddler is (according to him) moving forward two paces at a time, but he does this three times in a negative direction. If he were stepping backwards two paces at a time while facing forwards, that would be -2 times 3. Have a look at [Multiplying by Negatives ] for a longer description.
The Website
And that is why I have developed (Math is Fun , or "Maths is Fun" in British English), to be a place where mathematics can be explained in a more "user-friendly" manner.
And like all people who embark on explaining Science to the general public I must at times leave out details which would only confuse, but it can be very hard to know where to draw the line.
So please forgive me, fellow Wikipedians, when I over-simplify! And correct me gently, but do correct me!
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Test Area Stats
χ 2 = ∑ ( O − E ) 2 E
Test Area Taylor
f ( x ) = f ( a ) + f ′ ( a ) 1 ! ( x − a ) + f ″ ( a ) 2 ! ( x − a ) 2 + f ‴ ( a ) 3 ! ( x − a ) 3 + ⋯ .
e x = 1 + x + x 2 2 ! + x 3 3 ! + ⋯
e x = ∑ n = 0 ∞ x n n !
sin x = x − x 3 3 ! + x 5 5 ! − ⋯
sin x = ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! x 2 n + 1
cos x = 1 − x 2 2 ! + x 4 4 ! − ⋯
cos x = ∑ n = 0 ∞ ( − 1 ) n ( 2 n ) ! x 2 n
tan x = x + x 3 3 + 2 x 5 1 5 + ⋯ for | x | < π 2
tan x = ∑ n = 1 ∞ B 2 n ( − 4 ) n ( 1 − 4 n ) ( 2 n ) ! x 2 n − 1 for | x | < π 2
1 1 − x = 1 + x + x 2 + x 3 + ⋯ for | x | < 1
1 1 − x = ∑ n = 0 ∞ x n for | x | < 1
e x = ∑ n = 0 ∞ x n n ! = 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! + x 5 5 ! + ⋯
e i x = 1 + i x + ( i x ) 2 2 ! + ( i x ) 3 3 ! + ( i x ) 4 4 ! + ( i x ) 5 5 ! + ⋯
e i x = 1 + i x − x 2 2 ! − i x 3 3 ! + x 4 4 ! + i x 5 5 ! − ⋯
e i x = ( 1 − x 2 2 ! + x 4 4 ! − ⋯ ) + i ( x − x 3 3 ! + x 5 5 ! − ⋯ )
e i x = cos x + i sin x
Test Area Scratch
Help:Displaying_a_formula
r x y = n ∑ x i y i − ∑ x i ∑ y i n ∑ x i 2 − ( ∑ x i ) 2 n ∑ y i 2 − ( ∑ y i ) 2 .
f ( t a + ( 1 − t ) b ) ≥ t f ( a ) + ( 1 − t ) f ( b )
| A | = a ⋅ | e f h i | − b ⋅ | d f g i | + c ⋅ | d e g h |
| A | = a ⋅ | f g h j k l n o p | − b ⋅ | e g h i k l m o p | + c ⋅ | e f h i j l m n p | − d ⋅ | e f g i j k m n o |
Test Area Symbols
⇒ ⟺ ≈
Test Area Stats
r x y = ∑ i = 1 n ( x i − x ¯ ) ( y i − y ¯ ) ∑ i = 1 n ( x i − x ¯ ) 2 ∑ i = 1 n ( y i − y ¯ ) 2 ,
Variance: σ 2 = 2 0 6 2 + 7 6 2 + ( − 2 2 4 ) 2 + 3 6 2 + ( − 9 4 ) 2 5 = 4 2 , 4 3 6 + 5 , 7 7 6 + 5 0 , 1 7 6 + 1 , 2 9 6 + 8 , 8 3 6 5 = 1 0 8 , 5 2 0 5 = 2 1 , 7 0 4
Test Area Sigma
∑ f = 1 5 + 2 7 + 8 + 5 = 5 5
∑ f x = 1 5 × 1 + 2 7 × 2 + 8 × 3 + 5 × 4 = 1 1 3
x ¯ = ∑ f x ∑ f = 1 5 × 1 + 2 7 × 2 + 8 × 3 + 5 × 4 1 5 + 2 7 + 8 + 5 = 2 . 0 5 . . .
∑ k = m n c a k = c ∑ k = m n a k
∑ k = m n 6 k 2 = 6 ∑ k = m n k 2
∑ k = m n ( a k + b k ) = ∑ k = m n a k + ∑ k = m n b k
∑ k = m n ( a k − b k ) = ∑ k = m n a k − ∑ k = m n b k
∑ k = m n ( k + k 2 ) = ∑ k = m n k + ∑ k = m n k 2
∑ n = 1 4 ( 2 n + 1 ) = 3 + 5 + 7 + 9 = 2 4
∑ i = 2 1 4 ( $ 7 × 4 ( i − 1 ) + $ 1 1 × ( i − 2 ) 2 )
∑ i = 2 1 4 $ 7 × 4 ( i − 1 ) + ∑ i = 2 1 4 $ 1 1 × ( i − 2 ) 2
$ 7 × 4 ∑ i = 2 1 4 ( i − 1 ) + $ 1 1 × ∑ i = 2 1 4 ( i − 2 ) 2
$ 7 × 4 ∑ j = 1 1 3 j + $ 1 1 ∑ k = 1 1 2 k 2
$ 7 × 4 × 1 3 × 1 4 2 + $ 1 1 × 1 2 × 1 3 × 2 5 6
$ 7 × 4 × 9 1 + $ 1 1 × 6 5 0
$ 7 × 3 6 4 + $ 1 1 × 6 5 0
$ 2 5 4 8 + $ 7 1 5 0 = $ 9 6 9 8
Test Area Partial Sums
∑ k = 1 n 1 = n
∑ k = 1 n c = n c
∑ k = 1 n k = n ( n + 1 ) 2
∑ k = 1 n k 2 = n ( n + 1 ) ( 2 n + 1 ) 6
∑ k = 1 n k 3 = ( n ( n + 1 ) 2 ) 2
∑ k = 1 n k 3 = ( ∑ k = 1 n k ) 2
∑ k = 1 n ( 2 k − 1 ) = n 2
∑ k = 1 1 4 k 2 = 1 4 ( 1 4 + 1 ) ( 2 ⋅ 1 4 + 1 ) 6 = 1 0 1 5
∑ k = 0 n − 1 ( a + k d ) = n 2 ( 2 a + ( n − 1 ) d )
∑ k = 0 1 0 − 1 ( 1 + k ⋅ 3 ) = 1 0 2 ( 2 ⋅ 1 + ( 1 0 − 1 ) ⋅ 3 )
∑ k = 0 n − 1 ( a r k ) = a ( 1 − r n 1 − r )
∑ k = 0 4 − 1 ( 1 0 ⋅ 3 k ) = 1 0 ( 1 − 3 4 1 − 3 ) = 4 0 0
∑ k = 0 1 0 − 1 1 2 ( 1 2 ) k = 1 2 ( 1 − ( 1 2 ) 1 0 1 − 1 2 ) = 1 2 ( 1 − 1 1 0 2 4 1 2 ) = 1 − 1 1 0 2 4
∑ k = 0 6 4 − 1 ( 1 ⋅ 2 k ) = 1 ( 1 − 2 6 4 1 − 2 )
∑ n
∑ n = 1 4 n
∑ n = 1 4 n = 1 + 2 + 3 + 4 = 1 0
∑ n = 1 4 n 2 = 1 2 + 2 2 + 3 2 + 4 2 = 3 0
∑ i = 1 3 i ( i + 1 ) = 1 ⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 = 2 0
∑ i = 3 5 i i + 1 = 3 4 + 4 5 + 5 6
Test Area Binomial
( a + b ) n = ∑ k = 0 n ( n k ) a n − k b k
( n k ) ( 1 n ) k = n ! k ! ( n − k ) ! ⋅ 1 n k
∑ k = 0 ∞ 1 k ! = 1 0 ! + 1 1 ! + 1 2 ! + 1 3 ! + 1 4 ! + . . . = 1 + 1 + 1 2 + 1 6 + 1 2 4 + . . .
( a + b ) 3 = ∑ k = 0 3 ( 3 k ) a 3 − k b k = ( 3 0 ) a 3 − 0 b 0 + ( 3 1 ) a 3 − 1 b 1 + ( 3 2 ) a 3 − 2 b 2 + ( 3 3 ) a 3 − 3 b 3 = 1 ⋅ a 3 b 0 + 3 ⋅ a 2 b 1 + 3 ⋅ a 1 b 2 + 1 ⋅ a 0 b 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3
( 1 + 1 n ) n = ∑ k = 0 n ( n k ) 1 n − k ( 1 n ) k = ∑ k = 0 n ( n k ) ( 1 n ) k = ∑ k = 0 n n ! k ! ( n − k ) ! ⋅ 1 n k
( x + 5 ) 4 = ∑ k = 0 4 ( 4 k ) x 4 − k 5 k = ( 4 0 ) x 4 − 0 5 0 + ( 4 1 ) x 4 − 1 5 1 + ( 4 2 ) x 4 − 2 5 2 + ( 4 3 ) x 4 − 3 5 3 + ( 4 4 ) x 4 − 4 5 4 = 1 ⋅ x 4 5 0 + 4 ⋅ x 3 5 1 + 6 ⋅ x 2 5 2 + 4 ⋅ x 1 5 3 + 1 ⋅ x 0 5 4 = x 4 + 4 x 3 5 + 6 x 2 5 2 + 4 ⋅ 5 3 + 5 4
∑ k = 0 1 0 − 1 1 2 ( 1 2 ) k = 1 2 ( 1 − ( 1 2 ) 1 0 1 − 1 2 ) = 1 2 ( 1 − 1 1 0 2 4 1 2 ) = 1 − 1 1 0 2 4 = 0 . 9 9 9 0 2 3 4 3 7 5
Test Area Sigma 2
σ = 1 N ∑ i = 1 N ( x i − μ ) 2
s = 1 N − 1 ∑ i = 1 N ( x i − x ‾ ) 2
∑ k = 0 n − 1 ( a r k ) = a ( 1 − r n 1 − r )
∑ k = 0 ∞ ( a r k ) = a ( 1 1 − r )
∑ k = 0 ∞ ( 1 2 ⋅ ( 1 2 ) k ) = 1 2 ( 1 1 − 1 2 )
∑ k = 0 ∞ ( 1 2 ⋅ ( 1 2 ) k ) = 1 2 ( 1 1 − 1 2 )
0 . 9 9 9 . . . = 0 . 9 + 0 . 0 9 + 0 . 0 0 9 + . . . = 0 . 9 ⋅ 0 . 1 0 + 0 . 9 ⋅ 0 . 1 1 + 0 . 9 ⋅ 0 . 1 2 + . . . = ∑ k = 0 ∞ 0 . 9 ⋅ 0 . 1 k
∑ k = 0 ∞ 0 . 9 × 0 . 1 k = 0 . 9 ( 1 1 − 0 . 1 ) = 0 . 9 ( 1 0 . 9 ) = 1
Test Area Trig
a 2 − b 2 a
a 2 + b 2 a
a sin A = b sin B = c sin C
sin A a = sin B b = sin C c
c 2 = a 2 + b 2 − 2 a b cos ( C )
a 2 = b 2 + c 2 − 2 b c cos ( A )
b 2 = a 2 + c 2 − 2 a c cos ( B )
tan θ = sin θ cos θ
sin θ cos θ = O p p o s i t e / H y p o t e n u s e A d j a c e n t / H y p o t e n u s e = O p p o s i t e A d j a c e n t = tan θ
cot θ = cos θ sin θ
sin θ 2 = ± 1 − cos θ 2
cos θ 2 = ± 1 + cos θ 2
tan θ 2 = ± 1 − cos θ 1 + cos θ = sin θ 1 + cos θ = 1 − cos θ sin θ = csc θ − cot θ
cot θ 2 = ± 1 + cos θ 1 − cos θ = sin θ 1 − cos θ = 1 + cos θ sin θ = csc θ + cot θ
My Test Area Other
c = x 2 + y 2
c = ( − 2 ) 2 + 3 2 = 4 + 9 = 1 3
c = ( 3 − 9 ) 2 + ( 2 − 7 ) 2
c = ( − 6 ) 2 + ( − 5 ) 2 = 3 6 + 2 5 = 6 1 = 7 . 8 1 . . .
c = ( − 3 − 7 ) 2 + ( 5 − ( − 1 ) ) 2
c = ( − 1 0 ) 2 + ( 6 ) 2 = 1 0 0 + 3 6 = 1 3 6 = 1 1 . 6 6 . . .
c = ( 9 − 3 ) 2 + ( 7 − 2 ) 2
c = 6 2 + 5 2 = 3 6 + 2 5 = 6 1 = 7 . 8 1 . . .
c = ( x A − x B ) 2 + ( y A − y B ) 2
c = ( 9 − 4 ) 2 + ( 2 − 8 ) 2 + ( 7 − 1 0 ) 2 = 2 5 + 3 6 + 9 = 7 0 = 8 . 3 7 . . .
2 x 2 − 5 x − 1 x − 3 = 2 x + 1 + 2 x − 3
1 2
1 2 × 2 2 = 2 2
a m n = ( a n ) m
2 7 2 3 = ( 2 7 3 ) 2 = 3 2 = 9
4 6 3 = ( 4 ) 6 3 = 4 2 = 1 6
a m n = a m n
a n = a 1 n
2 3 3 = 2
− 2 3 3 = − 2
− 2 4 4 = | − 2 | = 2
5 4 = 6 2 5 s o 5 = 6 2 5 4
a b n = a n ⋅ b n
1 2 8 3 = 6 4 ⋅ 2 3 = 6 4 3 ⋅ 2 3 = 4 2 3
a b n = a n b n
1 6 4 3 = 1 3 6 4 3 = 1 4
a + b n ≠ a n + b n
a − b n ≠ a n − b n
a n + b n n ≠ a + b
a n n = a
a × a = a
a 3 × a 3 × a 3 = a
a n × a n × . . . × a n ⏟ n o f t h e m = a
Ellipse a and b
h = ( a − b ) 2 ( a + b ) 2
p = π ( a + b ) ( 1 + ∑ n = 1 ∞ ( 0 . 5 n ) 2 ⋅ h n )
p = π ( a + b ) ∑ n = 0 ∞ ( 0 . 5 n ) 2 h n
p = π ( a + b ) ( 1 + 1 4 h + 1 6 4 h 2 + 1 2 5 6 h 3 + . . . )
Ellipse perimeter, simple formula:
p ≈ 2 π a 2 + b 2 2
A better approximation by Ramanujan is:
p ≈ π [ 3 ( a + b ) − ( 3 a + b ) ( a + 3 b ) ]
p ≈ π ( a + b ) ( 1 + 3 h 1 0 + 4 − 3 h )
e = a 2 − b 2 a
p = 2 a π ( 1 − ∑ i = 1 ∞ ( 2 i ) ! 2 ( 2 i ⋅ i ! ) 4 ⋅ e 2 i 2 i − 1 )
p = 2 a π [ 1 − ( 1 2 ) 2 e 2 − ( 1 ⋅ 3 2 ⋅ 4 ) 2 e 4 3 − ( 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ) 2 e 6 5 − … ]
p = 2 a π [ 1 − ( 1 2 ) 2 e 2 − ( 1 ⋅ 3 2 ⋅ 4 ) 2 e 4 3 − … − ( 1 ⋅ 3 ⋅ 5 ⋅ … ⋅ ( 2 n − 1 ) 2 ⋅ 4 ⋅ 6 ⋅ … ⋅ 2 n ) 2 e 2 n 2 n − 1 − … ]
A x 2 + B x y + C y 2 + D x + E y + F = 0
Ellipse r and s
h = ( r − s ) 2 ( r + s ) 2
p = π ( r + s ) ( 1 + ∑ n = 1 ∞ ( 0 . 5 n ) 2 ⋅ h n )
p = π ( r + s ) ∑ n = 0 ∞ ( 0 . 5 n ) 2 h n
p = π ( r + s ) ( 1 + 1 4 h + 1 6 4 h 2 + 1 2 5 6 h 3 + . . . )
Ellipse perimeter, simple formula:
p ≈ 2 π r 2 + s 2 2
A better approximation by Ramanujan is:
p ≈ π [ 3 ( r + s ) − ( 3 r + s ) ( r + 3 s ) ]
ε = r 2 − s 2 r
p = 2 r π ( 1 − ∑ i = 1 ∞ ( 2 i ) ! 2 ( 2 i ⋅ i ! ) 4 ⋅ ε 2 i 2 i − 1 )
p = 2 r π [ 1 − ( 1 2 ) 2 ε 2 − ( 1 ⋅ 3 2 ⋅ 4 ) 2 ε 4 3 − ( 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ) 2 ε 6 5 − … ]
p = 2 r π [ 1 − ( 1 2 ) 2 ε 2 − ( 1 ⋅ 3 2 ⋅ 4 ) 2 ε 4 3 − … − ( 1 ⋅ 3 ⋅ 5 ⋅ … ⋅ ( 2 n − 1 ) 2 ⋅ 4 ⋅ 6 ⋅ … ⋅ 2 n ) 2 ε 2 n 2 n − 1 − … ]
A x 2 + B x y + C y 2 + D x + E y + F = 0
My Test Exponents
x 1 n = x n
2 7 1 3 = 2 7 3 = 3
x m n = x m n
6 2 5 n = 5
6 2 5 4 = 5
a n n = a
x m n = x m n = ( x n ) m
x m n = x m n = ( x n ) m
x 2 3 = x 2 3 = ( x 3 ) 2
x m n = x ( m × 1 n ) = ( x m ) 1 n = x m n
x m n = x ( 1 n × m ) = ( x 1 n ) m = ( x n ) m
My Test Area
∫ a b f ( x ) d x
2 x 2 + 5 x + 3 = 0
x 2 − 3 x = 0
5 x − 3 = 0
lim x → 0 sin ( x ) x = 1
1 − e 2
2 3 3 × 3 3 = 2 × 3 3 × 3 = 6 9
1 2 = 2 4 = 2 4 = 2 2
3 4 = 3 4 = 3 2
x 2 3 = x 2 3
1 0 1 0 1 0 0
1 0 1 0 1 0 1 0 0 0
n ! ( n − r ) ! × 1 r ! = n ! r ! ( n − r ) !
n ! r ! ( n − r ) ! = ( n r )
f ( k ; n , p ) = ( n k ) p k ( 1 − p ) n − k
for k = 0 , 1 , 2 , … , n and where
( n k ) = n ! k ! ( n − k ) !
f ( 3 ; 1 0 , 0 . 5 ) = ( 1 0 3 ) 0 . 5 3 ( 1 − 0 . 5 ) ( 1 0 − 3 ) = ( 1 0 3 ) 0 . 5 3 0 . 5 7
( 1 0 3 ) = 1 0 ! 3 ! ( 1 0 − 3 ) ! = 1 0 ! 3 ! 7 ! = 1 2 0
( 4 2 ) = 4 ! 2 ! ( 4 − 2 ) ! = 4 ! 2 ! 2 ! = 4 ⋅ 3 ⋅ 2 ⋅ 1 2 ⋅ 1 ⋅ 2 ⋅ 1 = 6
f ( 3 ; 1 0 , 0 . 5 ) = 1 2 0 × 0 . 5 3 0 . 5 7 = 0 . 1 1 7 1 8 7 5
P ( n , r ) = n P r = n P r = n ! ( n − r ) !
C ( n , r ) = n C r = n C r = ( n r ) = n ! r ! ( n − r ) !
P ( n , r ) = n ! ( n − r ) ! .
Test Area 2
i → × j → = k →
File:Hexadecimal multiplication table.svg A hexadecimal multiplication table
n C r = n ! ( n − r ) ! ( r ! )
∑ k = 1 ∞ 1 0 − k ! = 0.110001000000000000000001000...
0 < | x − p q | < 1 q n
A = s ( s − a ) ( s − b ) ( s − c )
C = c o s − 1 ( a 2 + b 2 − c 2 2 a b )
φ = 1 2 + 5 2 = 1 + 5 2
φ = 1 + 1 1 + 1 1 + 1 1 + ⋯ = 1 . 6 1 8 . . .
1 3 − 2
1 3 − 2 × 3 + 2 3 + 2 = 3 + 2 3 2 − ( 2 ) 2 = 3 + 2 7
2 − x 4 − x
2 − x 4 − x × 2 + x 2 + x = 2 2 − ( 2 ) 2 ( 4 − x ) ( 2 + x ) = ( 4 − x ) ( 4 − x ) ( 2 + x ) = 1 2 + x
Test Area Comb Perm
n r
n ! ( n − r ) !
n ! r ! ( n − r ) !
( r + n − 1 ) ! r ! ( n − 1 ) !
( r + n − 1 r ) = ( r + n − 1 ) ! r ! ( n − 1 ) !
( r + n − 1 r ) = ( r + n − 1 n − 1 ) = ( r + n − 1 ) ! r ! ( n − 1 ) !
Test Area Sets
f ( t ) = { $ 5 0 if t ≤ 6 $ 8 0 if t > 6 and t ≤ 1 5 $ 8 0 + $ 5 ( t − 1 5 ) if t > 1 5
f ( x ) = { x 2 if x < 2 6 if x = 2 1 0 − x if x > 2 and x ≤ 6 .
f ( x ) = | x | = { x , if x ≥ 0 − x , if x < 0 .
h ( x ) = { 2 , if x ≤ 1 x , if x > 1 .
1 x 1 y x 2 y x n y n y 1 n
e x l n ( y ) a x l o g a ( y )
s i n ( x ) a r c s i n ( y ) s i n − 1 ( y )
c o s ( x ) a r c c o s ( y ) c o s − 1 ( y )
t a n ( x ) a r c t a n ( y ) t a n − 1 ( y )
Help:Displaying_a_formula
f : ℕ → ℕ
f : { 1 , 2 , 3 , . . . } → { 1 , 2 , 3 , . . . }
f : ℝ → ℝ
f : x ↦ x 2
f ( x ) = x 2
From Set-builder notation
Examples:
∑ n = 1 4 ( 2 n + 1 )
-
Test Area Limits
lim x → 1 x 2 − 1 x − 1 = 2
lim x → 1 x 2 − 1 x − 1 = lim x → 1 ( x − 1 ) ( x + 1 ) x − 1 = lim x → 1 ( x + 1 )
lim x → 1 ( x + 1 ) = 1 + 1 = 2
x 2 − 1 x − 1 = ( x − 1 ) ( x + 1 ) x − 1 = x + 1
lim n → ∞ ( 1 + 1 n ) n = e
lim x → 1 0 x 2 = 5
lim x → 4 2 − x 4 − x
2 − x 4 − x × 2 + x 2 + x = 2 2 − ( x ) 2 ( 4 − x ) ( 2 + x ) = ( 4 − x ) ( 4 − x ) ( 2 + x ) = 1 2 + x
2 − x 4 − x × 2 + x 2 + x
2 2 − ( x ) 2 ( 4 − x ) ( 2 + x )
( 4 − x ) ( 4 − x ) ( 2 + x )
1 2 + x
lim x → 4 2 − x 4 − x = lim x → 4 1 2 + x = 1 2 + 4 = 1 4
Test Area Derivatives
2 x Δ x + Δ x 2 Δ x
lim Δ x → 0 2 x Δ x + Δ x 2 Δ x = lim Δ x → 0 2 x + Δ x
lim Δ x → 0 2 x + Δ x = 2 x
lim Δ x → 0 2 x Δ x + Δ x 2 Δ x = lim Δ x → 0 2 x Δ x Δ x = lim Δ x → 0 2 x = 2 x
lim Δ x → 0 3 x 2 Δ x + 3 x Δ x 2 + Δ x 3 Δ x = lim Δ x → 0 3 x 2 + 3 x Δ x + Δ x 2 = 3 x 2
f ′ ( x ) = lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x
Δ y Δ x = f ( x + Δ x ) − f ( x ) Δ x
d y d x = f ( x + d x ) − f ( x ) d x
d y d x
= f ( x + d x ) − f ( x ) d x
= ( x + d x ) 2 − x 2 d x
= x 2 + 2 x ⋅ d x + d x 2 − x 2 d x
= 2 x ⋅ d x + d x 2 d x
= 2 x + d x
= 2 x
Test Area Integrals
∫ 1 2 2 x d x = 2 2 − 1 2 = 3
∫ 0 . 5 1 c o s ( x ) d x = s i n ( 1 ) − s i n ( 0 . 5 ) = 0 . 8 4 1 . . . − 0 . 4 7 9 . . . = 0 . 3 6 2 . . .
∫ 1 3 c o s ( x ) d x = s i n ( 3 ) − s i n ( 1 ) = 0 . 1 4 1 . . . − 0 . 8 4 1 . . . = − 0 . 7 0 0 . . .
∫ 0 1 s i n ( x ) d x = − c o s ( 1 ) − ( − c o s ( 0 ) ) = − 0 . 5 4 0 . . . − ( − 1 ) = 0 . 4 6 0 . . .
∫ π 4 π 2 c o s ( x ) d x