# Definite integral and Barrow's rule

What is a definite integral? We will start from a function $$f (x)$$, which (bounded) values are known in a closed interval $$[a, b]$$.

Now, if we divide this interval into $$n$$ smaller intervals, and in each subinterval we draw a rectangle of equal height to the value of the function in the average point of the subinterval, we get what can be seen in the following graph:

Taking more subintervals (taking $$n$$ bigger), these will be narrower. Then, we define the definite integral of $$f(x)$$ between $$a$$ and $$b$$ as the sum of the areas of these rectangles, when the number of subintervals tends to infinite (that is to say, the width of the rectangles tends to zero). And we will write

$$\displaystyle \int_{x}^b f(x) \ dx$$$The indefinite integral verifies the following properties: • Linearity of the integrand: Given two functions $$f (x)$$ and $$g (x)$$ defined in the interval $$[a, b]$$, and being $$K$$ any constant, then $$\displaystyle \begin{array} {l}\int_{a}^b f(x) \pm g(x) \ dx= \int_{a}^b f(x) \ dx \pm \int_{a}^b g(x) \ dx \\ \int_{a}^b K \cdot f(x) \ dx= K \cdot \int_{a}^b f(x) \ dx \end{array}$$$

• Additive Property of the interval: If $$f(x)$$ is defined in $$[a, b]$$ and $$c$$ belongs to the interval $$[a, b]$$, then: $$\displaystyle \int _{a}^b f(x) \ dx = \int _{a}^c f(x) \ dx + \int _{c}^b f(x) \ dx$$$This property will be very useful to calculate integrals of continuous functions by pieces, as we will later see in some examples. From this property it is also possible to deduced that$$\displaystyle \int _{a}^b f(x) \ dx =0$$$

Once we know what a definite integral is, let's see how to calculate them:

## First Fundamental Theorem of Calculus and Barrow's rule

Let $$f(x)$$ be a function and $$F (x)$$ its primitive (or antiderivative, or indefinite integral). Then $$F(x)=\int_{a}^x f(x) \ dx +C$$. As $$F(a)=\int_{a}^a f(x) \ dx +C=0+C$$, we have that $$C=F(a)$$ and $$F(x)-F(a)=\int_{a}^x f(x) \ dx$$, in particular: $$F(x)=\int_{a}^b f(x) \ dx +=F(b)-F(a)$$$. That is to say that to calculate the definite integral of $$f(x)$$ in the interval $$[a, b]$$ is as easy as to find the primitive of $$f(x)$$, calculate its value in the ends of the interval and substract them. Let's see some examples: $$\displaystyle \int_{0}^1 x^2 \ dx=\Big[\frac{x^3}{3}\Big]_{0}^1=\frac{1}{3}-0=\frac{1}{3}$$$ where $$\dfrac{x^3}{3}$$ is the primitive function of $$x^2$$ and $$\Big[\dfrac{x^3}{3}\Big]$$ means to evaluate $$\dfrac{x^3}{3}$$ in $$1$$ and $$0$$ and substract the values.

$$\displaystyle \int_{0}^2 e^{3x} \ dx = \dfrac{1}{3} \int_{0}^2 3e^{3x} \ dx=\dfrac{1}{3}[e^{3x}]_{0}^2=\dfrac{1}{3}(e^{3\cdot 2}-e^{3 \cdot 0})=\dfrac{1}{3}(e^6-e^0)=\dfrac{1}{3}(e^6-1)$$$$$\displaystyle \int_{0}^{\frac{\pi}{4}} \tan x \ dx =\int_{0}^{\frac{\pi}{4}} \frac {\sin x}{\cos x} \ dx=-\Big[\ln (\cos x)\Big]_{0}^{\frac{\pi}{4}}=-\ln (\cos \frac{\pi}{4})+\ln (\cos 0)=$$$ $$=-\ln \frac{\sqrt{2}}{2}+\ln 1=\frac{1}{2}\ln 2$$\$