Integration

Integration Topic 7

Integration is an inverse process for differentiation. dy If = f(x) then ∫ f(x) dx = y dx

Integration

Integration symbol is ∫ For a function f(x), we can evaluate the integral as: ∫ f(x) dx = F(x) + c , where c is a constant. 1

2

Integration

Techniques of Integration

There are 2 types of integration: indefinite integral and definite integral

Rules of integration Substitution method Integration by parts

1. ∫ f(x) dx is indefinite integral b

2. ∫a f ( x) dx is definite integral that involves limit a and b, where a is called lower limit of integration and b is called upper limit of integration. 3

Rules of Integration

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Rules of Integration

Rule 1: ∫ a dx = ax + c where a and c is constant

Rule 2: n ∫x dx =

x n +1 + c, where n = integer and n ≠ -1 n +1

Example Example ∫ 4 dx = 4x + c 1 1 ∫ dx = x + c 3 3

∫x 3dx = = 5

x 3 +1 +c 3+1

x4 +c 4

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1

Rules of Integration Rule 3:

Rules of Integration Rule 4: ∫ [f(x) + g(x)]dx = ∫ f(x)dx + ∫ g(x)dx

axn +1 + c, ∫ax dx = a∫x dx = n +1 where n = integer and n ≠-1 n

n

Example

∫4x 5 dx = 4∫x 5 dx =

Example ∫ x3 + 2x2 dx = ∫ x3 dx + 2∫ x2 dx

5 +1

4x +c 5 +1

=

4x 6 2x 6 = +c = +c 6 3

x 4 2x3 + +c 4 3

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Rules of Integration

Rules of Integration

Rule 5: ∫ [f(x) - g(x)]dx = ∫ f(x)dx - ∫ g(x)dx

Rule 6: ∫ ex dx = ex + c Rule 7: 1 ∫ eax dx = e ax + c

Example ∫ x3 - 6x7 dx = ∫ x3 dx - 6∫ x7 dx = =

4

8

4

8

x 6x 4 8 x 3x 4 4

a

Example

+c +c

2x ∫e dx = 9

Rules of Integration Rule 8: Rule 9: Example ∫

1 2x e 2x e +c = +c 2 2

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Substitution Method For a function f(x) that hard to evaluate the integral using rules of integration, the integral can be evaluated using substitution method:

1 ∫ dx = ln | x | + c x ∫

8

1 1 dx = ln | ax + b | + c ax + b a

∫(f(u)

1 1 dx = ln | 2x + 3 | + c 2x + 3 2 11

du ) dx = ∫f(u) du dx 12

2

Substitution Method

Substitution Method 2

Example1: Solve ∫ 2x dx x +1 Solution:

u = x2 + 1 du = 2x dx du dx = 2x

substitute u = x2 + 1 and dx =

Example2: Solve ∫ x 1 - 2x 3 Solution:

u = 1 - 2x 3

du 2x

x du 1 1 1 1 = ∫ du = ∫ du = ln | u | + c ∫ u 2x 2u 2 u 2 1 = ln | x 2 + 1 | + c 2

du = -6x 2 dx du dx = - 6x 2

dx

substitute u = 1 - 2x3 and dx =

du

- 6x2

1 1 1 x2 du 1 =∫ du = ∫ ∫ du = ln | u | + c -6 -6 u u - 6x2 - 6u 1 = ln |1 - 2x3 | + c -6

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Substitution Method Example3: Solve Solution:

u = 1 + ex du = ex dx du dx = x e

Integration by parts

x 1 + e x dx ∫e

Integration by parts formula: If u = f(x) and v = g(x) then ∫ u dv = uv - ∫ v du how to choose u and dv choose dv first (dv must include dx) and u is the leftover part Example: ∫ xe2x dx

du substitute u = 1 + e x and dx = x e 1 du x 2 ∫e u x = ∫ u du = ∫u du e 3

=

2 3 u2 + c = u2 + c 3 3 2 3 2 = (1 + e x ) 2 + c 3

14

u

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Integration by parts

dv

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Integration by parts

Rules to choose u and dv

Example: Solve ∫ xe2x dx u=x Solution:

if function is combination of x and exponential, then u = x and dv = exponential. if function is combination of x and logarithmic, then u = logarithmic and dv = x

∫u dv = uv - ∫ v du 1 1 = x e2x - ∫ e2xdx 2 2 =

17

dv = e2xdx

du =1 dx du = dx

v = ∫ e2xdx 1 = e2x 2

xe2x 1 2x xe2x 1 1 2x - ∫e dx = - ( e )+c 2 2 2 2 2 =

xe2x e2x +c 2 4

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3

Properties of Definite Integration

Integration by parts Example: Solve ∫ 2x3 ln x dx u = ln x Solution: = ln x

x 4 x 4 dx -∫ 2 2 x

x4 1 3 = ln x - ∫x dx 2 2 = ln x

dv = 2x dx

du 1 = dx x dx du = x

∫u dv = uv - ∫ v du

a - ∫a f(x) dx = 0

3

x4 x4 x4 1 x4 + c = ln x +c 2 2 4 2 8

v = ∫ 2x 3 dx = 2 ∫ x 3 dx =2

x4 4

=

x4 2

3x3 = x3 ∫0 3x dx = 3 0 2

[ ]

+ ∫bc f(x) dx = ∫ac f(x) dx

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Definite Integration

Example: Solve ∫023x2dx Solution: 2

b - ∫a f(x) dx

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Definite Integration

2

a b - ∫b f(x) dx = - ∫a f(x) dx

Example: Solve Solution:

2 ∫2 5x + 2x dx 2

2 ∫2 5x + 2x dx = 0 because upper limit = lower limit 2

2 0

= 23 - 0 =8 21

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Definite Integration Example: Solve ∫02 x2 + 3x dx Solution: 2 ∫0 x + 3x dx = 2

x3 3x2 + 3 2

2

0

23 3(22 ) + = -0 3 2 8 12 26 = + = 3 2 3

23

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