Definite Integrals
Definite integrals are those integrals that have limits. They are particularly useful for finding areas beneath curves. Before exploring this page and the concept of Definite Integrals, read Integration Fundamentals. We recommend that you are familiar with the First Fundamental Theorem of Calculus.
The Fundamental Theorem of Calculus – Part 2
The Fundamental Theorem of Calculusย is split into two parts. The formal definitions are too advanced for this page but we will explore what the theorem states loosely. The second part of the theorem, we often call the Second Fundamental Theorem of Calculus, relates to definite integration. See An Introduction to Integrationย for the first part of the theorem.
Definite Integrals
Definite integrals are integrals that have scalar limits (single number limits). Limits appear in pairs – an upper/lower limit to the right of the top/bottom of the integral. Integrals with limits are written in the form as follows:
The Second Fundamental Theorem of Calculus states that we can evaluate this integral as the difference between the antiderivative (or indefinite integral) evaluated at the given limits:
.
Here, 
is the antiderivative of 
, that is to say 
. We find the antiderivative by integrating . Note that when evaluating definite integrals, it is not necessary to include a constant of integration as it always disappears when substituting in the limits – see Example 1. Remind yourself of what a constant of integration is here. Also note that 
is sometimes written as ![\left[F(x)\right]_a^b](https://studywell.com/wp-content/ql-cache/quicklatex.com-f105dee64f99894432adb757cb187a78_l3.png "Rendered by QuickLaTeX.com")
– see Example 2.
Note that the constant of integration always disappears in this way and so is typically left out when performing definite integration.
Finding an area beneath a curve using definite integrals
The integral expression above: , has a physical interpretation. It represents the area beneath the graph of 
, above the 
-axis and between the lines 
and 
as we can see in the diagram. Click here for a more formal explanation of the integral and how integral expressions represent areas. Note that if the curve is below the -axis, then the integral will be negative – we must be careful when interpreting ‘negative areas’.
Exam questions may expect you to use definite integration to find the area beneath a curve. It is important to remember that, in some questions, you may have to evaluate some areas separately if some of the curve is below the -axis – see Example 3. In other questions they may expect you to find the area between a curve and a straight line in a variety of different scenarios – see Example 4 for one such scenario.
In more advanced levels of mathematics, a question may ask you to find the area between two curves – see Example 5.
Examples
Evaluate 
Solution:
The antiderivative, more commonly known as the integral, is given by
According to the above, it follows that:
Note that the constant of integration always disappears in this way and so is typically left out when performing definite integration.
Evaluate 
Solution:
Integrating the integrand (the expression inside the integral) gives
It follows that
![\begin{array}{lll}\int_4^9\left(6x^{\frac{1}{2}}-5x^{\frac{3}{2}}\right), dx&=&\left[4x^{\frac{3}{2}}-2x^{\frac{5}{2}}\right]_4^9\\&=&\left(4(9)^{\frac{3}{2}}-2(9)^{\frac{5}{2}}\right)-\left(4(4)^{\frac{3}{2}}-2(4)^{\frac{5}{2}}\right) \\&=&\left(108-486\right)-\left(32-64\right)=-346\end{array}](https://studywell.com/wp-content/ql-cache/quicklatex.com-d4f3623a2e23d9d73510b5adb130b316_l3.png "Rendered by QuickLaTeX.com")
See more on indices if you require help with evaluating the individual terms.
The diagram shows the curve with equation 
. Find the exact area shaded shown in green.
Solution:
As we mention above, we can find the two areas separately as the area to the left of the 
-axis will be negative (this affects the result when integrating in one go). In order to know which limits to use, we need to know the roots of . Since is already factorised we can see that the roots are 
, 
and 
. Note that we also need to write as a polynomial so that we can integrate:
Hence, the integrals we require are:
![\begin{array}{lll} \int_0^2\left(-x^3+x^2+2x\right)dx&=&\left[-\frac{1}{4}x^4+\frac{1}{3}x^3+x^2\right]_0^2\\&=&\left(-\frac{1}{4}(2)^4+\frac{1}{3}(2)^3+(2)^2\right)-0\\&=&\frac{8}{3}\end{array}](https://studywell.com/wp-content/ql-cache/quicklatex.com-28a2e2e4e3144507776b04580678256a_l3.png "Rendered by QuickLaTeX.com")
and
![\begin{array}{lll} \int_{-1}^0\left(-x^3+x^2+2x\right)dx&=&\left[-\frac{1}{4}x^4+\frac{1}{3}x^3+x^2\right]_{-1}^0\\&=&0-\left(-\frac{1}{4}(-1)^4+\frac{1}{3}(-1)^3+(-1)^2\right)\\&=&-\frac{5}{12}\end{array}](https://studywell.com/wp-content/ql-cache/quicklatex.com-1cd32c67f22eca807090007df62be862_l3.png "Rendered by QuickLaTeX.com")
Hence, the total area (once making the latter integral positive) is 
square units exactly.
The diagram shows the curve 
and the straight line 
. Find the shaded area shown in purple to 3 decimal places.
Solution:
We find the area shown in purple by subtracting the area beneath the quadratic from the trapezium beneath the straight line. To find the area beneath the quadratic we use definite integrals but we must first identify the limits. Clearly, the lower limit is 0. The upper limit, however, we need to find by solving 
, that is, finding the -coordinate of the intersection of the two graphs:
We find the area shown in purple by subtracting the area beneath the quadratic from the trapezium beneath the straight line. To find the area beneath the quadratic we use definite integrals but we must first identify the limits. Clearly, the lower limit is 0. The upper limit, however, we need to find by solving , that is, finding the -coordinate of the intersection of the two graphs:
See solving quadratics. Taking the positive -coordinate, we have the area beneath the quadratic as the following definite integral:
![\int_0^{1.686} x^2,dx=\left[\frac{1}{3}x^3\right]_0^{1.686}=1.598…](https://studywell.com/wp-content/ql-cache/quicklatex.com-d639df60fde79c0bd98ab79df0af1fb0_l3.png "Rendered by QuickLaTeX.com")
to 3 decimal places. The area of the trapezium (half the sum of the parallel sides multiplied by the distance between them) is given by
to 3 decimal places. Hence, the area of the region shaded in green is 4.083-1.598=2.485 square units to 3 decimal places.
Find the area bounded by the curve 
and 
.
Solution:
We begin by sketching the region required. By factorising, we can see that the negative quadratic has roots at 
and 
. Also by factorising, the positive quadratic has roots at and 
.
We can already see that both curves have the point 
in common. To find the other point we solve 
, that is, 
. Factorising gives 
and so roots are at (as expected) and 
. To find the area of the region between the two curves we must first find the entire area beneath between and . Then we find the area beneath between and and subtract it from the first area:
![\begin{array}{lll}\int_\frac{1}{2}^3\left(9-x^2\right)dx-\int_\frac{1}{2}^3\left(x^2-7x+12\right)dx&=&\int_\frac{1}{2}^3\left(-2x^2+7x-3\right)dx\\&=&\left[-\frac{2}{3}x^3+\frac{7}{2}x^2-3x\right]_{\frac{1}{2}}^3\\&=&\left(-18+\frac{63}{2}-9\right)-\left(-\frac{1}{12}+\frac{7}{8}-\frac{3}{2}\right)\\&=&\frac{125}{24}\end{array}](https://studywell.com/wp-content/ql-cache/quicklatex.com-2a22ac8cab3b919e0401cd0ed35b565b_l3.png "Rendered by QuickLaTeX.com")
Videos
An integral with limits that involves an unknown constant.
Definite Integration involving a quadratic and a straight lines hence requiring knowledge of simultaneous equations as well as how to find complex areas beneath a curve.
How to find the area sandwiched between two quadratic curves.