Core 2: Integration

This section covers all the basic ideas required for integrating in AS-level 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 by-parts 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  units2. This means the area under the curve, between the x-coordinates 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 .