# 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 Calculus **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 Integration 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:

$\int_a^b f(x)\, dx$

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:

$

\int_a^b f(x) \, dx=F(b)-F(a)

$.

Here, $F(x)$ is the antiderivative of $f(x)$, that is to say $F'(x)=f(x)$. We find the antiderivative $F(x)$ by integrating $f(x)$. 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 $F(b)-F(a)$ is sometimes written as $\left[F(x)\right]_a^b$ – 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: $\int_a^b f(x)\, dx$, has a physical interpretation. It represents **the area beneath the graph of $y=f(x)$, above the $x$-axis and between the lines $x=a$ and $x=b$ **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 $x$-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 $x$-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.

## Questions by Topic

## Examples of Definite Integrals

### Example 1

Evaluate $\int_1^4 2x^3\, dx$

The antiderivative, more commonly known as the integral, is given by

$F(x)=\int f(x)\,dx=\int 2x^3\, dx=\frac{2}{4}x^4+c=\frac{1}{2}x^4+c$

According to the above, it follows that:

$\begin{array}{l}\int_1^4 2x^3\, dx\\=F(4)-F(1)\\=\left(\frac{1}{2}(4)^4+c\right)-\left(\frac{1}{2}(1)^4+c\right)\\=128+c-\frac{1}{2}-c\\=127\frac{1}{2}\end{array}$

Note that the constant of integration always disappears in this way and so is typically left out when performing definite integration.

### Example 2

Evaluate $\int_4^9\left(6x^{\frac{1}{2}}-5x^{\frac{3}{2}}\right)\, dx$

Integrating the integrand (the expression inside the integral) gives

$\begin{array}{l}F(x)=\int\left(6x^{\frac{1}{2}}-5x^{\frac{3}{2}}\right)\, dx\\=\frac{6}{\frac{3}{2}}x^{\frac{3}{2}}-\frac{5}{\frac{5}{2}}x^{\frac{5}{2}}+c=4x^{\frac{3}{2}}-2x^{\frac{5}{2}}+c\end{array}$

It follows that

$\begin{array}{l}\int_4^9\left(6x^{\frac{1}{2}}-5x^{\frac{3}{2}}\right)\, dx=\left[4x^{\frac{3}{2}}-2x^{\frac{5}{2}}\right]_4^9\\=\left(4(9)^{\frac{3}{2}}-2(9)^{\frac{5}{2}}\right)-\left(4(4)^{\frac{3}{2}}-2(4)^{\frac{5}{2}}\right) \\=\left(108-486\right)-\left(32-64\right)=-346\end{array}$

See more on indices if you require help with evaluating the individual terms.

### Example 3

The diagram shows the curve with equation $y=-x(x-2)(x+1)$. Find the exact area shaded shown in green.

As we mention above, we can find the two areas separately as the area to the left of the $y$-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 $y$. Since $y$ is already factorised we can see that the roots are $x=-1$, $x=0$ and $x=2$. Note that we also need to write $y$ as a polynomial so that we can integrate:

$y=-x(x-2)(x+1)=-x(x^2-x-2)=-x^3+x^2+2x$

Hence, the integrals we require are:

$\begin{array}{l} \int_0^2\left(-x^3+x^2+2x\right)dx=\left[-\frac{1}{4}x^4+\frac{1}{3}x^3+x^2\right]_0^2\\=\left(-\frac{1}{4}(2)^4+\frac{1}{3}(2)^3+(2)^2\right)-0=\frac{8}{3}\end{array}$

and

$\begin{array}{l} \int_{-1}^0\left(-x^3+x^2+2x\right)dx=\left[-\frac{1}{4}x^4+\frac{1}{3}x^3+x^2\right]_{-1}^0\\=0-\left(-\frac{1}{4}(-1)^4+\frac{1}{3}(-1)^3+(-1)^2\right)=-\frac{5}{12}\end{array}$

Hence, the total area (once making the latter integral positive) is $\frac{8}{3}+\frac{5}{12}=\frac{37}{12}$ square units exactly.

### Example 4

The diagram shows the curve $y=x^2$ and the straight line $y=0.5x+2$. Find the shaded area shown in purple to 3 decimal places.

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 $x^2=0.5x+2$, that is, finding the $x$-coordinate of the intersection of the two graphs:

$x^2-0.5x-2=0\hspace{15pt}\Longrightarrow\hspace{15pt}x=-1.186…, 1.686…. $

See solving quadratics. Taking the positive $x$-coordinate, we have the area beneath the quadratic as the following definite integral:

$\int_0^{1.686} x^2\,dx=\left[\frac{1}{3}x^3\right]_0^{1.686}=1.598…$

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

$0.5\times(2+0.5\times 1.686+2)\times1.686=4.083…$

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.

### Example 5

(Advanced Maths such as Year 2 A-Level)

Find the area bounded by the curve $y=9-x^2$ and $y=x^2-7x+12$.

We. begin by sketching the region required. By factorising, we can see that the negative quadratic $y=9-x^2$ has roots at $x=3$ and $x=-3$. Also by factorising, the positive quadratic $y=x^2-7x+12$ has roots at $x=3$ and $x=4$.

We can already see that both curves have the point $(3,0)$ in common. To find the other point we solve $9-x^2=x^2-7x+12$, that is, $2x^2-7x+3=0$. Factorising gives $(2x-1)(x-3)=0$ and so roots are at $x=3$ (as expected) and $x=\frac{1}{2}$. To find the area of the region between the two curves we must first find the entire area beneath $y=9-x^2$ between $x=\frac{1}{2}$ and $x=3$. Then we find the area beneath $y=x^2-7x+12$ between $x=\frac{1}{2}$ and $x=3$ and subtract it from the first area:

$\begin{array}{l}&&\int_\frac{1}{2}^3\left(9-x^2\right)dx-\int_\frac{1}{2}^3\left(x^2-7x+12\right)dx\\&=&\int_\frac{1}{2}^3\left(-2x^2+7x-3\right)dx\\&=&\left[-\frac{2}{3}x^3+\frac{7}{2}x^2-3x\right]_{\frac{1}{2}}^3\\&=&\left(-18+\frac{63}{2}-9\right)-\left(-\frac{1}{12}+\frac{7}{8}-\frac{3}{2}\right)\\&=&\frac{125}{24}\end{array}$