Definite integrals

Area under the Curve and the Fundamental Theorem of Integral Calculus

Look at the curve $y = f\left( x \right)$ in the following image:

Area bounded by curve $y = f\left( x \right)$, axis $X$, line $x=a$, and line $x=x$ or area $AA'P'P$ defined by:

\[L\left( x \right) = \int\limits_a^x {f\left( x \right)} dx\]

Now suppose that $x$ changes to $\left( {x + \Delta x} \right)$, then the new area $\left( {AA'Q'Q} \right)$ changes to $L \left( {x + \Delta x} \right)$, so that the increase in area of $\left( {PP'Q'Q} \right)$ is determined by $L\left( {x + \Delta x} \right ) - L\left( x \right)$. By looking at the picture above, the following relationship is obtained:

$\begin{array}{l} \text{area of PP'Q'R} < \text{area of PP'Q'Q} < \text{area of SP'Q'Q}\\ \Leftrightarrow f\left( x \right).\Delta x < L\left( {x + \Delta x} \right) - L\left( x \right) < f\left( {x + \Delta x} \right).\Delta x\\ \Leftrightarrow f\left( x \right) < \frac{{L\left( {x + \Delta x} \right) - L\left( x \right)}}{{\Delta x}} < f\left( {x + \Delta x} \right) \end{array}$

For ${\Delta x}$ close to $0$, we get:

$\begin{array}{l} f\left( x \right) \le \mathop {\lim }\limits_{\Delta x \to 0} \frac{{L\left( {x + \Delta x} \right) - L\left( x \right)}}{{\Delta x}} \le \mathop {\lim }\limits_{\Delta x \to 0} f\left( {x + \Delta x} \right)\\ \mathop {\lim }\limits_{\Delta x \to 0} \frac{{L\left( {x + \Delta x} \right) - L\left( x \right)}}{{\Delta x}} = f\left( x \right)\\ \Leftrightarrow \frac{{dL\left( x \right)}}{{dx}} = f\left( x \right)\\ \Leftrightarrow dL\left( x \right) = f\left( x \right)dx \end{array}$

Using integral operations on each side of the equation above, we get:

$\begin{array}{l} \int {dL\left( x \right)} = \int\limits_a^x {f\left( x \right)dx} \\ \Leftrightarrow L\left( x \right) = \int\limits_a^x {f\left( x \right)dx} = F\left( x \right) + C \end{array}$

where $F\left( x \right)$ is the anti-differential of ${f\left( x \right)}$ which is $F'\left( x \right) = f\left( x \right )$.

From the relationship $L\left( x \right) = \int\limits_a^x {f\left( x \right)dx} = F\left( x \right) + C$ we can specify the following: :

• For $x=a$ get:
$\begin{array}{l} L\left( a \right) = \int\limits_a^a {f\left( x \right)dx} = F\left( a \right) + C = 0\\ \Leftrightarrow C = - F\left( a \right) \end{array}$

So that $L\left( x \right)$ can be written as:

$L\left( x \right) = F\left( x \right) - F\left( a \right) = \int\limits_a^x {f\left( x \right)dx} $
• For $x=b$ get:
$L\left( b \right) = \int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right)$

Based on the above equation, we get the relationship $\int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right)$. This relationship is known as the Basic Theorem of Integral Calculus.

Basic Theorem of Integral Calculus

Suppose the curve ${f\left( x \right)}$ is continuous in a closed interval $\left[ {a,b} \right]$. The area of $L$ bounded by the curve $y = f\left( x \right)$, axis $X$, line $x=a$, and line $x=b$ is determined by the formula:

\[L = \int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right)\]

where $F\left( x \right)$ is the anti-differential of ${f\left( x \right)}$ which is $F'\left( x \right) = f\left( x \right )$.

Writing $F\left( b \right) - F\left( a \right)$ can be abbreviated using the special notation $\left[ {F\left( x \right)} \right]_a^b$ , so that the Fundamental Theorem of Integral Calculus can be written as follows:

\[\int\limits_a^b {f\left( x \right)dx} = \left[ {F\left( x \right)} \right]_a^b = F\left( b \right) - F\left( a \right)\]

Definite Integral Properties

Suppose that $f\left( x \right)$ and $g\left( x \right)$ are continuous functions defined on closed intervals respectively $\left[ {a,b} \right]$ , then the general properties of certain integrals can be presented in the following summary form:

1. $\int\limits_a^a {f\left( x \right)} dx = 0$
2. $\int\limits_a^b {f\left( x \right)dx = - } \int\limits_b^a {f\left( x \right)dx} $
3. $\int\limits_a^b {kf\left( x \right)dx = k} \int\limits_a^b {f\left( x \right)dx} $
4. $\int\limits_a^b {\left\{ {f\left( x \right) \pm g\left( x \right)} \right\}} dx = \int\limits_a^b {f \left( x \right)} dx \pm \int\limits_a^b {g\left( x \right)} dx$
5. $\int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx = } } \int\limits_a^b {f\left ( x \right)dx,\left( {a < c < b} \right)} $
6. If $f\left( x \right) \ge 0$ in the interval $a \le x \le b$, then $\int\limits_a^b {f\left( x \right)dx \ge 0 } $.
7. If $f\left( x \right) \le 0$ in the interval $a \le x \le b$, then $\int\limits_a^b {f\left( x \right)dx \le 0 } $.

Problems Example

Find the following definite integral:

1. $\int\limits_1^2 {4dx} $
2. $\int\limits_0^1 {\left( {{x^2} - 3x + 4} \right)} dx$
3. $\int\limits_1^4 {\sqrt x } dx$
4. $\int\limits_0^\pi {\sin xdx} $
5. $\int\limits_0^{\frac{\pi }{2}} {\left( {{x^2} - \cos x} \right)dx} $

Answer

1. $\int\limits_1^2 {4dx} = \left[ {4x} \right]_1^2 = \left( {4.2} \right) - \left( {4.1} \right) = 8 - 4 = 4$
$\begin{array}{l} \int\limits_0^1 {\left( {{x^2} - 3x + 4} \right)} dx = \left[ {\frac{1}{3}{x^3} - \frac{3} {2}{x^2} + 4x} \right]_0^1\\ = \left( {\frac{1}{3}{1^3} - \frac{3}{2}{1^2} + 4.1} \right) - \left( {\frac{1}{3 }{0^3} - \frac{3}{2}{0^2} + 4.0} \right)\\ = \left( {\frac{1}{3} - \frac{3}{2} + 4} \right) - \left( 0 \right)\\ = \frac{2}{6} - \frac{9}{6} + \frac{{24}}{6}\\ = \frac{{17}}{6} = 2\frac{5}{6} \end{array}$
$\begin{array}{l} \int\limits_1^4 {\sqrt x } dx = \int\limits_1^4 {{x^{\frac{1}{2}}}} dx\\ = \left[ {\frac{1}{{\frac{1}{2} + 1}}{x^{\frac{1}{2} + 1}}} \right]_1^4\\ = \left[ {\frac{1}{{\frac{3}{2}}}{x^{\frac{3}{2}}}} \right]_1^4\\ = \left[ {\frac{2}{3}x\sqrt x } \right]_1^4\\ = \left( {\frac{2}{3}.4\sqrt 4 } \right) - \left( {\frac{2}{3}.1\sqrt 1 } \right)\\ = \left( {\frac{2}{3}.8} \right) - \left( {\frac{2}{3}.1} \right)\\ = \frac{{16}}{3} - \frac{2}{3}\\ = \frac{{14}}{3}\\ = 4\frac{2}{3} \end{array}$
$\begin{array}{l} \int\limits_0^\pi {\sin xdx} = \left[ { - \cos x} \right]_0^\pi \\ = \left( { - \cos \pi } \right) - \left( { - \cos 0} \right)\\ = \left( 1 \right) - \left( { - 1} \right)\\ = 2 \end{array}$
$\begin{array}{l} \int\limits_0^{\frac{\pi }{2}} {\left( {{x^2} - \cos x} \right)dx} = \left[ {\frac{1}{3}{ x^3} - \sin x} \right]_0^{\frac{\pi }{2}}\\ = \left( {\frac{1}{3}.{{\left( {\frac{\pi }{2}} \right)}^3} - \sin \left( {\frac{\pi } {2}} \right)} \right) - \left( {\frac{1}{3}{{.0}^3} - \sin 0} \right)\\ = \left( {\frac{1}{3}.\frac{{{\pi ^3}}}{8} - 1} \right) - \left( {\frac{1}{3}.0 - 0} \right)\\ = \frac{{{\pi ^3}}}{{24}} - 1\\ = \frac{{{\pi ^3} - 24}}{{24}} \end{array}$

Thus the discussion about certain integrals and examples. Hope it's useful