The word calculus comes from a latin word meaning small stones and is like understanding something by looking at small pieces. It can be used to find areas, volumes, central points and many useful things. Integrals can be of the type : single integral, double integral, multiple integrals. Generally these are used to find surface area and the volume of geometric solids. Integrals together with derivatives are the fundamental objects of calculus. Integral calculus is also known as antiderivative and primitive.  Integrals are commonly of two types : Definite integrals and the indefinite integrals .

## Formulae

List below are some of the very common formulas used in integral calculus.

1) $\int x^{n}dx$ = $\frac{x^{n+1}}{n+1}$ + $C$

2) $\int$ $\frac{1}{x}$ $dx$ = ln |x| + $C$

3) $\int$ $\frac{1}{\sqrt{x}}$ $dx$ = 2$\sqrt{x}$ + $C$

4) $\int$ ln $x dx$ = x ln x - x + $C$

5) $\int$ x$^{n}$dx = $\frac{x^{n+1}}{n+1}$ + $C$

6) $\int$ f (g(x)) g'(x) $dx$ = F(g(x))+$C$

If u = g(x), then
du = g'(x)dx & $\int$ f(u)du = F(u) + $C$

7) $\int$ u dv = uv - $\int$ v du

8) $\int_{a}^{b}$u dv = uv]$_{a}^{b}$ - $\int_{a}^{b}$ v du

9) $\int_{a}^{b}$f(x)g'(x)dx = f(x)g(x)]$_{a}^{b}$ - $\int_{a}^{b}$g(x)f'(x)dx

## Applications of Integral Calculus

Some cases where integral calculus can be applied in real life:

Graph visualization:
Using calculus it is easy to graph any function or equation. One can easily find the minimum and maximum values, also where it increases and decreases without even graphing a point.

Finding area of any shape:
It would be almost impossible to find the area of any geometrical shape if there was no calculus.

Finding average of a function:
While trying to find the path of the airplane calculus comes handy! Using calculus one can easily find average cruising altitude, acceleration and velocity. In the same way this can also be used for bus, car etc.,

Given below are some problems:
Example 1 : If the marginal revenue is 3x$^{2}$ - 4x + 3, find the total revenue and average revenue where x is the output. Write the demand function.

Solution : Marginal revenue : 3x$^{2}$ - 4x + 3

Total revenue [3 $\int$ x$^{2}$ - 4 $\int$ x + 3] dx

Average Revenue = $\frac{Total\ Revenue}{x}$

$\frac{3x^{3}}{3}$ - $\frac{4x^{2}}{2}$ + 3x

x$^{3}$ - 2x$^{2}$ + 3x

Average Revenue = Demand function = $\frac{x^{3}-2x^{2}+3x}{x}$

= x$^{2}$ - 2x + 3

Example  2 : The marginal cost function of a firm is 150 - 10x + 0.2x$^{2}$  where x is the output. Find the total cost function. If the total cost is Rs. 750 What is the average cost?

Solution : Marginal cost = 150 - 10x + 0.2x$^{2}$
Fixed cost = 750

Total cost = 150x - 10 $\frac{x^{2}}{2}$ + 0.2 $\frac{x^{3}}{3}$ + 750

Average cost = $\frac{Total\ Cost}{Q}$

= $\frac{150-10.\frac{x^2}{2} + 0.2 \frac{x^{3}}{3}+750} {x}$

= 150 - 5x + $\frac{0.2x^{2}}{3}$ + $\frac{750}{x}$

## Examples

Examples of integral calculus are given below:

Example 1 : Solve $\int$ $\frac{log x}{x}$ $dx$

Solution : $\int$ $\frac{log x}{x}$ $dx$ = $\int$ t dt where t = log x and $\frac{1}{x}$ $dx$ = $dt$

= $\frac{t^{2}}{2}$ + $C$

= $\frac{(logx)^{2}}{2}$ + $C$

Example 2 : Solve $\int$ $\frac{x^{e-1}+e^{x-1}}{x^{e}+ e^{x}}$

Solution : Put x$^{e}$+ e$^{x}$ = 1

then, (ex$^{e-1}+e^{x}$) $dx$ = $dt$

i.e, e[x$^{e-1}+ e^{x-1}$] $dx$ = $dt$

$\int$ $\frac{x^{e-1}+ e^{x-1}}{x^{e}+ e^{x}}$ $dx$

= $\frac{1}{e}$ $\int$ $\frac{dt}{t}$

= $\frac{1}{e}$ log t + c

= $\frac{1}{e}$ $log (x^{e}+e^{x})$ + c

Example 3 : $\int_{0}^{1}xe^{x^{2}}$dx

Solution : Put x$^{2}$ = t => 2x $dx$ = $dt$

or  xdx = $\frac{dt}{2}$

When x = 0 => t = 0
When x = 1 => t = 1

Now $\int_{0}^{1}e^{t}.\frac{dt}{2}$

= $\frac{1}{2}$ $\int_{0}^{1}e^{t}$ $dt$

t = 1,

$\frac{1}{2}$ $e^{t}]_{0}^{1}$

= $\frac{1}{2}$ $[e^{1}-e^{0}]$

= $\frac{1}{2}$ $[e-1]$

Example 4 : Find the area between the curves y$^{2}$ = 4x and x$^{2}$= 4y

Solution : First find the values of x by solving both the equations $(\frac{x^2}{4})^2$  = 4x

x$^{4}$ = 64x

x$^{4}$ - 64x = 0

x(x$^{3}$ - 64)= 0

x = 0, x  = 4

To find the area between given curves we need to subtract the region bounded by second curve from the first curve.

=> $\int_{0}^{4}\sqrt{4}\sqrt{x}dx$ - $\int_{0}^{4}$$\frac{x^{2}}{4}$ $dx$

= 2 $\times$ $\frac{2}{3}$ $\times$ $x^\frac{3}{2}]_{0}^{4}$  - $\frac{x^{3}}{12}$ $]_{0}^{4}$

= $\frac{4}{3}(4^{\frac{3}{2}})-\frac{1}{12}(4^{3}$)

= $\frac{32}{3}$ - $\frac{16}{3}$

= $\frac{16}{3}$