Browse Questions

# Evaluate the definite integral as limits of sums$\int\limits_1^4(x^2-x)dx$

Toolbox:
• $\int\limits_a^b f(x)dx=lim_{h->0}h[f(a)+f(a+h)+...f(a+(n-1)h)]$
• where $h=\large\frac{b-a}{n}$
• (ii)$\sum (n-1)=\frac{n(n-1)}{2} \qquad \sum(n-1)^2=\large\frac{n(n-1)(2n-1)}{6}$
Given $I=\int \limits_1^4(x^2-x)dx$

$f(x)=(x^2-x)$

Hence $I=\int \limits_1^4(x^2-x)dx=$

$lim_{h->0}h[f(a)+f(a+h)+...f(a+(n-1)h)]$

Here $a=1\;and\;b=4,therefore\; h=\frac{b-a}{n}=\frac{4-1}{n}=\frac{3}{n}$

f(a)=(1-1)=0

$f(a+h)=[(1+h)^2-(1+h)]$

$=1^2+2h+h^2-1-h$

$=(h+h^2)$

$f(a+2h)=[(1+2h)^2-(1+2h)]$

$=1+4h+4h^2-1-2h$

$=2h+4h^2$

$f(a+(n-1)h)=[(1+(n-1)h)^2-(1+(n-1)h)]$

$=1+(n-1)^2 h^2+2(n-1)h-1-(n-1)h$

$=(n-1)^2 h^2+(n-1)h$

Now substituting the above values we get

$\int\limits_1^4 (x^2-x))dx=lim_{h->0}h[0+(h+h^2)+(2h+4h^2+.....(n-1)^2h^2+(n-1)h]$

We can write the above function as

$lim_{h->0}h[h+2h+3h+......(n-1)h+h^2+4h^2+.....(n-1)^2 h^2]$

On seperating the common factor,

$=lim_{h->0}h[h[(1+2+3+....(n-1)]+h^2(1^2+2^2+.....(n-1)^2]$

Now substitute $h=\frac{3}{n},\sum(n-1)=\frac{n(n-1)}{2}\; and\; \sum(n-1)^2=\frac{n(n-1)(n-2)}{6}$

As $h \to 0 \qquad n \to \infty$

Hence $\int \limits_1^4 (x^2-x)dx=\frac{3}{n} \lim _{n \to \infty}\bigg[\large\frac{3}{n}\frac{n(n-1)}{2}+\frac{9}{n^2}\frac{n(n-1)(2n-1)} {6}\bigg]$

$\frac{3}{n} \lim _{n \to \infty}\bigg[\large\frac{3}{2}(n-1)+\frac{9}{n}\frac{n(n-1)(2n-1)}{6}\bigg]$

$=\lim _{n \to \infty}\large\frac{9}{2}(1-\frac{1}{n})+\frac{9}{2}(1-\frac{1}{n})(2-\frac{1}{n})$

Applying limits we get

$\large\frac{9}{2}+\frac{9}{2}(2)=\frac{9}{2}+9$

$=\large\frac{27}{2}$