Ucale

# Integration as the limit of a sum

In this article we shall consider integration as the limit of the sum of certain number of terms when the numbers of the terms tends to infinity and each term tends to zero. As a matter of fact the summation aspect of definite integrals is more fundamental and it was invented as far before the differentiation was known.

Let f(x) be a continuous real valued function defined on the closed interval [a,b] which is divided into n equal parts each of width h by intersecting (n-1)points a+h, a+2h, a+3h,……,a+(n+1)  between a and b as shown in the figure. Then, nh =b-a or $\displaystyle h=\frac { b-a }{ n }$

Let $\displaystyle { S }_{ n }$ the sum of the areas of n rectangles shown in figure then $\displaystyle { S }_{ n }=hf\left( a \right) +h.f\left( a+h \right) +h.f\left( a+2h \right) +....+h.f\left( a+\left( n-1 \right) h \right) \\ \Rightarrow { S }_{ n }=h\left[ f\left( a \right) +f\left( a+h \right) +f\left( a+2h \right) +....+f\left( a+\left( n-1 \right) h \right) \right]$

Clearly, $\displaystyle { S }_{ n }$ denotes the areas which is close to the area of the region bounded the curve y=f(x), x-axis and the ordinates x=a,x=b It is evident that if n will decrease. Consequently $\displaystyle { S }_{ n }$ gives closer approximation to the area enclosed by the curve y=f(x), x-axis and the ordinates x=a, x=b

Thus $\displaystyle \lim _{ n\rightarrow \infty }{ { S }_{ n } }$ gives the area of the region bounded by the four curves y=f(x), y=0 (x-axis), x=a and x=b. It can be proved that limit exists for all continuous functions defined on closed integral [a,b] and is defined as the  definite integral of f(x) over [a,b] $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx=\lim _{ n\longrightarrow \infty }{ { S }_{ n } }$ $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx=\lim _{ n\longrightarrow \infty }{ h\left[ f\left( a \right) +f\left( a+h \right) +f\left( a+2h \right) +....+f\left( a+\left( n-1 \right) h \right) \right] } \\where\quad h=\frac { b-a }{ n } \\ or\quad \int _{ a }^{ b }{ f\left( x \right) } dx=\lim _{ h\longrightarrow 0 }{ h\left[ f\left( a \right) +f\left( a+h \right) +f\left( a+2h \right) +....+f\left( a+\left( n-1 \right) h \right) \right] } \qquad \left[ \because n\longrightarrow \infty \Leftrightarrow h\longrightarrow 0 \right] \qquad ......(i)$

The process of evaluating a definite integral integral by using the above definition is called integration from first principles or integration by ab-initio method or integration as the limit of a sum.

Remark

In finding $\displaystyle { S }_{ n }$ in the above article,  we have taken the left end points of the sub intervals. We can also take right end points of the sub intervals throughout to obtain: $\displaystyle or\quad \int _{ a }^{ b }{ f\left( x \right) } dx=\lim _{ n\longrightarrow \infty }{ h\left[ f\left( a \right) +f\left( a+h \right) +f\left( a+2h \right) +....+f\left( a+\left( n-1 \right) h \right) \right] } \\where\quad h=\frac { b-a }{ n }$

It can be proved that this formula and the formula (i) give the same limit.

The following result will be helpful in evaluating definite integrals as limit of sums. $\displaystyle \left( i \right) \quad 1+2+3+.....+\left( n-1 \right) =\frac { n\left( n-1 \right) }{ 2 }$ $\displaystyle \left( ii \right) \quad { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+.....+{ \left( n-1 \right) }^{ 2 }=\frac { n\left( n-1 \right) \left( 2n-1 \right) }{ 6 }$ $\displaystyle \left( iii \right) \quad { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+.....+{ \left( n-1 \right) }^{ 3 }={ \left\{ \frac { n\left( n-1 \right) }{ 2 } \right\} }^{ 2 }$ $\displaystyle \left( iv \right) \quad a+ar+a{ r }^{ 2 }+.....+a{ r }^{ n-1 }=a{ \left\{ \frac { { r }^{ n }-1 }{ r-1 } \right\} },\quad r\neq 1$ $\displaystyle \left( v \right) \quad \sin { a } +\sin { \left( a+h \right) } +\sin { \left( a+2h \right) } +.....+\sin { \left( a+\left( n-1 \right) h \right) } \\=\frac { \sin { \left\{ a+\left( \frac { n-1 }{ 2 } \right) h \right\} } \sin { \left( \frac { nh }{ 2 } \right) } }{ \sin { \left( \frac { h }{ 2 } \right) } }$ $\displaystyle \left( vi \right) \quad \cos { a } +\cos { \left( a+h \right) } +\cos { \left( a+2h \right) } +.....+\cos { \left( a+\left( n-1 \right) h \right) } \\ =\frac { \cos { \left\{ a+\left( \frac { n-1 }{ 2 } \right) h \right\} } \sin { \left( \frac { nh }{ 2 } \right) } }{ \sin { \left( \frac { h }{ 2 } \right) } }$

### Example $\displaystyle \int _{ 0 }^{ 2 }{ \left( x+4 \right) } dx$

Solution:

We have $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx=\lim _{ h\longrightarrow 0 }{ h\left[ f\left( a \right) +f\left( a+h \right) +f\left( a+2h \right) ....+f\left( a+\left( n-1 \right) h \right) \right] } dx\\ where\quad h=\frac { b-a }{ n } .Here\quad a=0,b=2,f\left( x \right)=x+4\quad and\quad h=\frac { 2-0 }{ n } =\frac { 2 }{ n }$ $\displaystyle \therefore \int _{ 0 }^{ 2 }{ \left( x+4 \right) } dx=\lim _{ h\longrightarrow 0 }{ { h\left[ f\left( 0 \right) +f\left( 0+h \right) +f\left( 0+2h \right) ....+f\left( 0+\left( n-1 \right) h \right) \right] } } \\ =\lim _{ h\longrightarrow 0 }{ { h\left[ f\left( 0+4 \right) +f\left( h+4 \right) +f\left( 2h+4 \right) ....+f\left( \left( n-1 \right) h+4 \right) \right] } }$ $\displaystyle =\lim _{ h\longrightarrow 0 }{ { h\left[ 4n+h\frac { n\left( n-1 \right) }{ 2 } \right] } } \\ =\lim _{ n\longrightarrow \infty }{ \frac { 2 }{ h } \left[ 4n+h\frac { n\left( n-1 \right) }{ 2 } \right] } \qquad \left[ \because h=\frac { 2 }{ n } \quad and\quad h\longrightarrow 0\Rightarrow n\longrightarrow \infty \right]$ $\displaystyle =\lim _{ n\longrightarrow \infty }{ \left[ 8+2\left( 1-\frac { 1 }{ n } \right) \right] } \\ =8+2\left( 1-0 \right) =10$

April 18, 2019
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