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Core 2: Integration
This section covers all the basic
ideas required for integrating in ASlevel Mathematics. Negative areas and the
trapezium rule will be added to integration part 2, which is currently being
written.
This section also does not cover
skills such as integration by substitution, integration byparts and functions
other than polynomials. To revise these, read the CORE 3 revision guide on
integration.
The fundamental formula for integrating a polynomial is
given by the general formula: The answer is called the integrand 
Think about this as: “Increase the power by 1 and then divide by the new power” The is important and is called the constant
of integration Integration is used for lots of purposes, though the most regular
use you will have will be to find area under curves and to solve differential
equations 
Question 
Answer 








Harder powers and more complicated
expressions 




In this example observe how the new power just reciprocates to
give the multiplier of x. 

To find the new power, add the denominator on to the
numerator and then multiply by the reciprocal of the new power 
Using the integrand to solve
differential equations This uses the principle that integration is the reverse
of differentiation 

Solve the differential equation: 
This questions means that an expression was differentiated to
give an answer of . To find the original expression we need to reverse this by
integrating. This is called a general solution. This is an unknown constant, which may have been
there, but which would have disappeared upon differentiating. 
Finding the constant of integration 
You need to know some extra information, often called boundary
or initial conditions depending on their form 
Find the equation of the curve, which passes though the point and which has the gradient function: Therefore, 
First of all integrate To find ,
substitute the given coordinate into the expression. This is called a particular solution 
Using the integrand to find areas The integrand is sometimes known as the area function 

Find the area enclosed by the curve: and the lines and We need to do: We now substitute in the values: and and find the difference. The area is units^{2}. 
This means the area under the curve, between the
xcoordinates 1 and 2. The 1 and 2 are called limits We have integrated. You do NOT need to worry about the .
Observe that we have written the limits adjacent to the square brackets. The brackets are essential, especially if you have negative
numbers. 
A more concise example: 

Find the area under the curve between the values and . 
