Integral Maths - Definite Integrals

Niamh Ryan
Note by Niamh Ryan, updated more than 1 year ago
Niamh Ryan
Created by Niamh Ryan almost 4 years ago


In this our final study note on our series on Differentiation and Integration, we look at definite integrals. Using graphs we demonstrate the key points that you need to understand to master the topic. Sample equation questions are then provided to allow you to check your learning progress.

Resource summary

Page 1

What is a definite integral?

So far we have just considered integration as the reverse of differentiation. However, integration does have uses other than just reversing differentiation. The integral of a function between two points represents the area enclosed by the x axis, the curve described by the function and the two chosen points. For example, consider the function drawn below. Integrating f(x) between a and b will give us the value of the shaded area.

Page 2

Example 1

Example: Find the integral of the following function between the given limits: \[ \int^{4}_{3} 3x^2+4 dx\]   Answer: First evaluate the indefinite integral. \[ \int 3x^2+4 dx = \frac{3x^3}3 + 4x = x^3 +4+C\] To calculate the definite integral, sub the upper and lower limits into the integrand, then find the difference between these two expressions. \[ \int^{4}_{3} 3x^2+4 dx = \Big[ x^3+4 \Big]^4_3 = \Big(4^3+4 \Big) - \Big(3^  3+4 \Big) \]\[=64+4-27-4=64-27=37\] Notice that there is no need for a constant of when calculating the definite integral. The constants would cancel when you subtract the expressions for the upper and lower limits.

Page 3

Negative answers

Sometimes when calculating a definite integral, you may end up with an answer that is a negative number.  A negative area represents an area below the x axis, as illustrated in the diagram shown below.   If a question specifically asks you to calculate the area, make sure to find the absolute value of the area rather than the sum total of the positive and negative areas.

Page 4

Example 3

  Example: Find the area enclosed by the function \(f(x) = 4x^3\) between \(x=-2 \text{ and } x=2\)\\   Answer: If we integrate this as usual, we will get answer that doesn't make sense. \[ \int ^2_{-2} 4x^3dx=\Big[x^4 \Big]^2_{-2}=2^4-(-2)^4=0\]Clearly , as you can see in the graph below, the area enclosed by the function isn't zero. The integration has given us an answer of zero because it counts area below the x-axis as negative area. To find the absolute value of the area, we must calculate the area enclosed by the function from \(x=-2 \text{ to } x=0\) and \(x=0 \text{ to } x=-2\) separately.  \[\int ^2_{0} 4x^3dx=\Big[x^4 \Big]^2_{0}=2^4-(0)^4=16\] \[\int ^0_{-2} 4x^3dx=\Big[x^4 \Big]^0_{-2}=0-(-2)^4=-16\] Now, to find the area, we sum the absolute value of these two integrands. \[Area = 16+\lvert -16\rvert =32\]

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