Curve is an object similar to a line but not be straight or we can say generalization of a line. We already know about a straight line. If we have a line segment, then we can find the length of the segment easily by simply measuring the length using a ruler or a scale or by placing the line segment on a grid. This is shown in the figure below

Line Segment

From the above figure we can easily state that the length of the given line segment is 5 units

In another case if we have a line segment joining two points, we can find its length using the distance formula as follows:

Length of a Line Segment

Here, we have the coordinates of the two points. So the distance between them can be found as follows:

$D$ = $\sqrt{((x_{2}\ -\ x_{1})^{2}\ +\ (y_{2}\ -\ y_{1})^{2})}$

$D$ = $\sqrt{((2\ -\ (-1))^{2}\ +\ (5-(-3))^{2})}$

$D$ = $\sqrt{(3^{2}\ +\ 8^{2})}$

$D$ = $\sqrt{73}$

Now we shall learn to do the same for a curve as well.

Like a line, a curve is also a plane figure. The difference being that, a curve need not be a straight line. A curve may deviate from being a straight line gradually. It may be curved for the entire or a part of the plane figure. 

Technically, a curve can be defined as a continuous function, $f : R \rightarrow R$. For a plane curve, the domain and co domain both would be real numbers. The curve need not be a single variable curve. It may also be a bi variable curve. A curve in 3D space would be of the form $z$ = $f(x,y)$.
The length of a curve in 2D plane can be found as follows. Let us begin by assuming that the curve follows the function, $y$ = $f(x)$. This function is defined over the interval $[a,b]$. We shall assume that the function is differentiable over the interval $(a,b)$. Then, the length of the above curve can be calculated using the formula:

$L$ = $\int_a^b\ \sqrt{(1\ +\ (\frac{dy}{dx})^{2})}$  $dx$

If we have $x$ as a function of $y$, then the function would be of the form $x$ = $g(y)$ over the interval $[c,d]$. This function is assumed to be differentiable over the interval $(c,d)$. The length of such a curve can be found using the following formula:

$L$ = $\int_c^d\ \sqrt{(1\ +\ (\frac{dx}{dy})^{2})}$  $dy$

If the equation of the curve is known to us in parametric form, then the length of the curve can be found as below. 

Let’s say that the parametric equation of a curve is known to us as $x$ = $f(t)$  and $y$ = $g(t)$ over the interval $[a,b]$. Also the function is differentiable over the interval $(a,b)$. Then the length of the curve in the given interval can be given by the formula:

$L$ = $\int_a^b$ $\sqrt{((\frac{dx}{dt})^{2}\ +\ (\frac{dy}{dt})^{2})}$ $dt$

There are many types of curves known in Math. Some of them are listed below:


No. Curve General Equation Properties
1 Parabola $y$ = $a(x-h)^{2}\ +\ k$ Conic section curve
2 Circle $(x-h)^{2}\ +\ (y-k)^{2}$ = $r^{2}$ Conic section curve
3 Ellipse $\frac{(x-h)^{2}}{a^{2}}$ + $\frac{ (y-k)^{2}}{b^{2}}$ = $1$ Conic section curve
4 Hyperbola $\frac{(x-h)^{2}}{a^{2}}$ - $\frac{(y-k)^{2}}{b^{2}}$ = $1$ or $\frac{(y-k)^{2}}{b^{2}}$ - $\frac{(x-h)^{2}}{a^{2}}$ = $1$ Conic section curve
5 Logarithmic $y$ = $log_{b}\ (ax)$
6 Exponential $y$ = $ab^{x}$
7 Reciprocal $y$ = $\frac{k}{(ax\ +\ b)}$
8 Quadratic $y$ = $ax^{2}\ + \ bx\ +\ c$ Same as parabola. A polynomial curve.
9 Cubic $y$ = $ax^{3}\ +\ bx^{2}\ +\ cx\ +\ d$ Also a polynomial curve
10 Polynomial curves $y$ = $a_{1}\ x^{n}\ +\ a_{2}\ x^{(n-1)}\ +\ a_{3}\ x^{(n-2)}\ +…+\ a_{(n-1)}\ x\ +\ a_{n}$ Nth degree polynomial curve
11 Trigonometric $y$ = $aSin(bx+c)\ +\ d$ The sin can be replaced with any other trigonometric function.
12 Cardioid $r$ = $a(1-cos(t))$ Polar curve
13 Cycloid of Ceva $r$ = $1\ +\ 2cos(2t)$ Polar curve
14 Limacon $r$ = $b\ +\ acos(t)$ Polar curve

These are a few of the common curves. Besides these there may be hundreds of other curves. 
Consider a curve $y$ = $f(x)$ over the interval $[a,b]$. The area enclosed by this curve, the lines $x$ = $a$, $x$ = $b$ and the $x$ axis, would be called the area under the curve within the given interval. Graphically it can be represented as follows:

Area Under a Curve

There are various method that can be used for calculating this area. 

This can be seen in the following figure:


Application on Area under Curve

Note that the area under the curve $f(x)$ is split to rectangles. We can find the area of each of those rectangles and add them up to get an estimate of the area under the curve. 

In the above case, left sum has been used. The same can be calculated using the right sum or also the midpoint sum. The figures for the same are as below.

Right Sum of a Curve Area

As the number of rectangles increase the estimate of area gets more and more accurate. 
To find the exact area under a curve, this is the method that has to be used. For a curve, $y$ = $f(x)$ over the interval $[a,b]$, the area under the curve is given by the integral:

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

This is derived from the fact that if the number of rectangles in the above method is infinite with the width of each rectangle equal to $dx$, then the sum would be limit of a sequence. The limit of a sequence can be depicted as a definite integral.
Example 1:

Find the length of the curve, $f(x)$ = $ln(sec(x))$ over the interval $[0,\frac{pi}{4}]$.

Solution: 

Length of curve = $L$ = $\int_a^b \sqrt{(1\ +\ (\frac{dy}{dx})^{2})}$ $dx$

$y$ = $ln(sec(x))$

$\frac{dy}{dx}$ = $\frac{(sec(x)\ tan(x))}{sec(x)}$ = $tan(x)$

$(\frac{dy}{dx})^{2}$ = $tan^{2}(x)$

$1$ + $(\frac{dy}{dx})^{2}$ = $1\ +\ tan^{2}(x)$

$\sqrt{(1\ +\ (\frac{dy}{dx})^{2})}$ = $\sqrt{(1\ +\ tan^{2}(x))}$ = $\sqrt{(sec^{2}(x))}$ = $sec(x)$

$\therefore\ L$ = $\int_0^(\frac{p}{4})$ $sec(x)\ dx$ = $ln|sec(x)\ +\ tan(x)$ $|_0^\frac{\pi}{4}$

$L$ = $Ln(\sqrt2\ +\ 1) \leftarrow Answer!$
Example 2:

Find the area under the curve, $f(x)$ = $\frac{1}{x}$ within the interval $[1,4]$.

Solution:

Area under curve = $A$ = $\int_a^b\ f(x)dx$

$A$ = $\int_1^4\ \frac{1}{x}$ $dx$

$A$ = $ln(x)\ |_1^4$

$A$ = $Ln(4)\ -\ Ln(1)$ = $Ln(4)\ \leftarrow\ Answer!$