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May 21, 2023
The reverse process of differentiation
Differentiation
The process of finding the rate of change of one variable compared to another. Useful when the rate of change...
6/11/2023
, where a derived equation is restored to its original equation. There is a loss when differentiating an equation (i.e., constants are stripped away), which is why integration is only able to add an arbitrary constant C C C or K K K to the restored equation.
Has a notation of ∫ f ( x ) d x \int f(x) \space dx ∫ f ( x ) d x , read as “the integral of f ( x ) f(x) f ( x ) with respect to x x x ”.
f ( x ) f(x) f ( x ) is known as the integrand;F ( x ) F(x) F ( x ) is known as anti-derivative of f ( x ) f(x) f ( x ) C C C an arbitrary constant of integration.# RulesAs with differentiation, integration has several rules that are the opposite of their counterparts with differentiation. These rules include the:
power rule; reciprocal rule. sum and difference rule; constant multiple rule; and product rule. # Power rule∫ x n d x = x n + 1 n + 1 + C
\int x^n \space dx = \frac {x^{n + 1}} {n + 1} + C
∫ x n d x = n + 1 x n + 1 + C
If the integral equation is in the form ( a x + b ) n (ax + b)^n ( a x + b ) n , the following formula can be used:
∫ ( a x + b ) n = ( a x + b ) n + 1 a ( n + 1 )
\int (ax + b)^n = \frac {(ax + b)^{n + 1}} {a(n + 1)}
∫ ( a x + b ) n = a ( n + 1 ) ( a x + b ) n + 1
# Reciprocal ruleWhen the exponent of x x x is − 1 -1 − 1 , it’s not possible to use the power rule as it’ll leave with a fraction dividing by zero. In this case, this rule is used instead:
∫ 1 x d x = ∫ x − 1 d x = l n ∣ x ∣
\int \frac 1 x \space dx = \int x^{-1} \space dx = ln \mid x \mid
∫ x 1 d x = ∫ x − 1 d x = l n ∣ x ∣
# Sum and difference rule∫ [ f ( x ) + g ( x ) ] d x = ∫ f ( x ) d x + ∫ g ( x ) d x
\int [f(x) + g(x)] \space dx = \int f(x) \space dx + \int g(x) \space dx
∫ [ f ( x ) + g ( x )] d x = ∫ f ( x ) d x + ∫ g ( x ) d x
∫ [ f ( x ) − g ( x ) ] d x = ∫ f ( x ) d x − ∫ g ( x ) d x
\int [f(x) - g(x)] \space dx = \int f(x) \space dx - \int g(x) \space dx
∫ [ f ( x ) − g ( x )] d x = ∫ f ( x ) d x − ∫ g ( x ) d x
# Constant multiple rule∫ k f ( x ) d x = k ∫ f ( x ) d x
\int kf(x) \space dx = k \int f(x) \space dx
∫ k f ( x ) d x = k ∫ f ( x ) d x