Comprehensive Guide to Integral Calculus

The ideas that led to integral calculus were introduced long time ago. However, these ideas were not developed systematically. The determinations of volume and area were found in Egyptian literature of 1820 BC. These methods did not have the major principles of differentiation. The Greeks in 408-355BC employed a method of inscribing polygons inside a shape to determine areas and volumes. The Greek mathematician, Archimedes of 287-212 BC employed the discovery methods to find the area of a circle. This was later re-discovered by Liu about 3 AD (Liu et al., 1996). 

The basic ideas and results

Differential calculus is the study of the calculating the derivative (gradient function) and its applications. This is referred to as differentiation.

i) The gradient linear functions

The equation of a straight line is given as can be written as y = ax + c,

Where y and x are  variables,

m is the gradient of the line,  

c is the y-intercept.{\displaystyle m={\frac {\text{rise}}{\text{run}}}={\frac {{\text{change in }}y}{{\text{change in }}x}}={\frac {\Delta y}{\Delta x}}.}                                                                 

The slope of the straight line (m) is constant.

ii) The equation of a curve (y=xn)

 The gradient of a curve at a point is given by the gradient of a tangent to the curve at that point. The gradient of a curve thus changes from point to point.

Suppose we want to get the gradient of the curve y=x2 at a general point (x, x2). Take points (x, x2) and [x+∆x, (x+∆x)2 ] where ∆x is a small increment in x. The gradient of the secants can be worked out as tabulated in table 1 below.

{\displaystyle m={\frac {f(a+h)-f(a)}{(a+h)-a}}={\frac {f(a+h)-f(a)}{h}}.}Table 1: Gradients of secants to the curve y=x2

Secant between points

∆x

∆y=f(x+∆x)-f(x)

Gradient of secant=∆y/∆x

x=1,       x=3

2

8

4

x=1,       x=2

1

3

3

x=1,       x=1.5

0.5

1.25

2.5

x=1 ,      x=1.1

0.1

0.21

2.1

x=1 ,      x=1.001

0.001

0.002001

2.001

From table 1, as → Δx approaches 0, the gradients of the secants approach 2.

∆y/∆x = [x+∆x)2 – x2] /∆x

            =[(x2+2x∆x +(∆x)2 – x2] / ∆x]

            = 2x +∆x           

As ∆x → to 0, the secant becomes the tangent to the curve and the gradient function becomes                        

dy/dx =   lim  ∆x → 0       (2x+∆x) = 2x   

The above gradient function can be used to get the gradient of the curve y=x2 at any point,

At x=1,      dy/dx = 2

     x=2,     dy/dx = 4

     x=3      dy/dx = 6

From the above example the gradient functions of curves are tabulated as below;

function

Gradient function

 y=x2

2x

y=x3

3x2

y=x4

4x3

y=x5

5x4

In general the gradient function o f y=xn

is given by nxn-1 where n is a positive integer.

{\displaystyle m={\frac {f(a+h)-f(a)}{(a+h)-a}}={\frac {f(a+h)-f(a)}{h}}.}Applications

Differentiation is applied in many mathematical problems as explained below;

a)Velocity and acceleration

The displacement s meters, covered by a moving particle after time, t seconds, is given by s=2t3 +4t2-8t+3.Find

i) The velocity when t=2 seconds                        

ii) Instant at which the particle is at rest.

Solution

Velocity is obtained by differentiating displacement with respect to time.

Hence ds/dt =2*3t2 +4*2t – 8

                      =6 t2+8t-8

On substituting t=2 on the above equation,

v=ds/dt=6*22+8*2 -8

         =24+16-8

          =32 m/s

ii) the particle is at rest when v=o

    hence 6 t2+8t-8=0

             2(3t2+4t-4) =0

             2(3t-2)(t+2) =0

             (3t-2)(t+2) =0

             (3t-2)=0 or (t+2) =0

            3t=2 or t= -2

             t= 2/3 or t=-2

            t=2/3 seconds

b) Maxima and minima

A farmer has 100 m of wire-mesh to fence a rectangular piece of land. What is the greatest area he can fence with this wire-mesh?

Solution

Let length, width, and perimeter be L, W, and P, respectively,

Hence P = 2(L+ W) … (5)

Then    100 =2(L +W)

Upon division by two, we get

50=L+W

W= 50-L ……………………………… (6)

The area of the land is  given by

A=L*W

    =L(50-L)

     = 50L-L2 ……………………………..(7)

Differentiation of (7) gives

dA/dL = 50- 2L=0 (For a maxima or minima da/dL = 0) ………..(8)

Simplification of (8) yields

50= 2L

Hence L = 25 m

From (6), w=50-25=25 m

Therefore maximum area, A =25m*25m

                                               = 625 m2

Work Cited

Lopez, César Pérez. “Integration and Applications.” MATLAB Differential and Integral             Calculus, 2014, pp. 129–162., doi:10.1007/978-1-4842-0304-0_6.