Integral Maths - Definite Integrals

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.

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.

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.

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. \[Area = 16+\lvert -16\rvert =32\]