Understanding The Integral Calculus
over the past few weeks i have received many emails from various people asking me to solve some questions and abstract problems on the integral calculus,and so many questions on this topics are been posted on many mathematics groups on facebook.
The Integral calculus is one aspect of calculus that seems to be a problem to many students both at the high school and undergraduate levels.Although Students seem to have a better understanding of the differential calculus,so in order to elaborate this aspect that have been a problem to many students i will be writing my next few topics mostly on integral calculus.
The integral calculus as we know it, is divided into two parts,we have the "definite integral" and the "indefinite integral".
I will begin my discussion with the definite integral.
The Integral calculus is one aspect of calculus that seems to be a problem to many students both at the high school and undergraduate levels.Although Students seem to have a better understanding of the differential calculus,so in order to elaborate this aspect that have been a problem to many students i will be writing my next few topics mostly on integral calculus.
The integral calculus as we know it, is divided into two parts,we have the "definite integral" and the "indefinite integral".
I will begin my discussion with the definite integral.
Definite Integral
The definite integral of a function $f(x)>0$ from $x=a$ to $x=b$ for $b>a$, is defined as the area bounded by the vertical lines $x=a$ and $x=b$ which is the x-axis, and the curve is given by $y=f(x)$, Now this area under the curve is obtained by a limit. first the area is approximated by a sum of rectangle areas.
second, the integral is defined to be the limit of the rectangle areas as the width of the individual rectangle goes to zero and the number of rectangle goes to infinity.hence this resulting infinite sum is called a Riemann sum which is defined by
\begin{equation*}
\int^b_a{f(x)}dx=\lim_{h\rightarrow{0}}\sum^N_{n=1}{f(a+(n-1)h})h
\end{equation*}
where $N=\frac{(b-a)}{h}$ is the number of terms in the sum.
the Riemann sum definition is extended to all values of a and b and for all values of positive and negative . Accordingly
\begin{align*}
\int^a_b{f(x)dx}=-\int^b_a{f(x)}dx\\
\int^b_a(-f(x))dx=\int^b_a{f(x)dx}\\
\text{Also,if}\ a<b<c\ \text{then}\\
\int^c_a{f(x)}dx=\int^b_a{f(x)}dx+\int^c_b{f(x)}dx
\end{align*}
which states that when $f(x)>0$ the total area equals the sum of its parts.
In a nutshell definition, the definite integral can be said to be the Riemann sum of any function.
The definite integral of a function $f(x)>0$ from $x=a$ to $x=b$ for $b>a$, is defined as the area bounded by the vertical lines $x=a$ and $x=b$ which is the x-axis, and the curve is given by $y=f(x)$, Now this area under the curve is obtained by a limit. first the area is approximated by a sum of rectangle areas.
second, the integral is defined to be the limit of the rectangle areas as the width of the individual rectangle goes to zero and the number of rectangle goes to infinity.hence this resulting infinite sum is called a Riemann sum which is defined by
\begin{equation*}
\int^b_a{f(x)}dx=\lim_{h\rightarrow{0}}\sum^N_{n=1}{f(a+(n-1)h})h
\end{equation*}
where $N=\frac{(b-a)}{h}$ is the number of terms in the sum.
the Riemann sum definition is extended to all values of a and b and for all values of positive and negative . Accordingly
\begin{align*}
\int^a_b{f(x)dx}=-\int^b_a{f(x)}dx\\
\int^b_a(-f(x))dx=\int^b_a{f(x)dx}\\
\text{Also,if}\ a<b<c\ \text{then}\\
\int^c_a{f(x)}dx=\int^b_a{f(x)}dx+\int^c_b{f(x)}dx
\end{align*}
which states that when $f(x)>0$ the total area equals the sum of its parts.
In a nutshell definition, the definite integral can be said to be the Riemann sum of any function.
Indefinite Integral
Now we move on to the indefinite integral, Normally an indefinite integral is defined by the expression \begin{align*}
\int{f(x)dx}=F(x)
\end{align*}
$F(x)$ is an anti-derivative of $f(x)$, in fact we can say that the indefinite integral is the reverse of the derivative of a function.
\subsection*{Indefinite integrals of the elementary functions}
From our known derivatives of elementary functions, we can determine some simple indefinite integrals. the power rule
\begin{equation*}
\int{x^n}dx=\frac{x^{n+1}}{n+1}+c,
\end{equation*}
this is the case when $n\neq{-1}$ and c is the constant of integration.
Now lets see when $n=-1$ and $x$ is positive,we have
\begin{equation*}
\int{x^{-1}}dx=\int\frac{1}{x}dx=\ln{x}+c
\end{equation*}
and that's what we got a log function.
if we set $x$ to be negative, then here the chain rule of differentiation is applicable.
\begin{equation*}
\frac{d}{dx}\ln(-x)=\frac{1}{x}
\end{equation*}
therefore since
\begin{equation*}
\vert{x}\vert{=}
\begin{cases}
-x & \text{if $x<0$}\\
x, & \text{if $x>0$}
\end{cases}
\end{equation*}
Now we can generalise our indefinite integral to strictly positive or strictly negative x:
\begin{equation*}
\int{\frac{1}{x}dx}=\ln|x|+c
\end{equation*}
In next posts i will be discussing on the Riemann integral and Riemann sums in full details.
Now we move on to the indefinite integral, Normally an indefinite integral is defined by the expression \begin{align*}
\int{f(x)dx}=F(x)
\end{align*}
$F(x)$ is an anti-derivative of $f(x)$, in fact we can say that the indefinite integral is the reverse of the derivative of a function.
\subsection*{Indefinite integrals of the elementary functions}
From our known derivatives of elementary functions, we can determine some simple indefinite integrals. the power rule
\begin{equation*}
\int{x^n}dx=\frac{x^{n+1}}{n+1}+c,
\end{equation*}
this is the case when $n\neq{-1}$ and c is the constant of integration.
Now lets see when $n=-1$ and $x$ is positive,we have
\begin{equation*}
\int{x^{-1}}dx=\int\frac{1}{x}dx=\ln{x}+c
\end{equation*}
and that's what we got a log function.
if we set $x$ to be negative, then here the chain rule of differentiation is applicable.
\begin{equation*}
\frac{d}{dx}\ln(-x)=\frac{1}{x}
\end{equation*}
therefore since
\begin{equation*}
\vert{x}\vert{=}
\begin{cases}
-x & \text{if $x<0$}\\
x, & \text{if $x>0$}
\end{cases}
\end{equation*}
Now we can generalise our indefinite integral to strictly positive or strictly negative x:
\begin{equation*}
\int{\frac{1}{x}dx}=\ln|x|+c
\end{equation*}
In next posts i will be discussing on the Riemann integral and Riemann sums in full details.
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