Maths Integrals

We know that  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}=\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[f\left(a\right)+f\left(a+h\right)+f\left(a+2h\right)+......+f\left(a+\left(n-1\right)h\right)\right]$

where nh = b - a

Here, a = 1, b = 4, nh = 3 and f(x) = x² - xf(x) =  x² - x 

$\begin{array}{l}\therefore {\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2+x\right)dx}=\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[0+h+h+2h+4h^2+......+\left(n-1\right)h+{\left(n-1\right)}^2{h}^2\right]\\ =\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[h\left(1+2+3+......+\left(n-1\right)\right)+h^2\left(1^2+2^2+......{\left(n-1\right)}^2\right)\right]\\ =\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[4nh+4hh\frac{n\left(n-1\right)}{2}+hhh\frac{n\left(n-1\right)\left(2n-1\right)}{6}\right]\\ =\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[h\left(1+2+3+......+\left(n-1\right)\right)+h^2\left(1^2+2^2+......{\left(n-1\right)}^2\right)\right]\\ =\underset{h\to 0}{lim}h\left[3\frac{\left(3-h\right)}{2}+\frac{3\left(3-h\right)\left(2.3-h\right)}{6}\right]\\ =\left[3\frac{\left(3-0\right)}{2}+\frac{3\left(3-0\right)\left(6-0\right)}{6}\right]\\ =\left[\frac{9}{2}+9\right]=\frac{27}{2}\end{array}$

We know that  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}=\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[f\left(a\right)+f\left(a+h\right)+f\left(a+2h\right)+......+f\left(a+\left(n-1\right)h\right)\right]$

where nh = b - a

Here, a = 2, b = 3, nh = 1 and f(x) = x²

$\begin{array}{l}\therefore {\displaystyle \underset{a}{\overset{b}{\int }}{x}^{2}dx}=\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[4+\left(4+4h+h^2\right)+\left(4+8h+2^2{h}^2\right)+......+\left(4+4\left(n-1\right)h+{\left(n-1\right)}^2{h}^2\right)\right]\\ =\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[4n+4h+\left(1+2+3+......+\left(n-1\right)\right)+h^2\left(1^2+2^2+......{\left(n-1\right)}^2\right)\right]\\ =\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[4nh+4hh\frac{n\left(n-1\right)}{2}+hhh\frac{n\left(n-1\right)\left(2n-1\right)}{6}\right]\\ =\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[4nh+4hh\frac{\left(nh-h\right)}{2}+\frac{nh\left(nh-h\right)\left(2nh-h\right)}{6}\right]\end{array}$

$\begin{array}{l}=\underset{h\to 0}{lim}h\left[4+2\left(1-h\right)+\frac{\left(1-h\right)\left(2-h\right)}{6}\right]\\ =\left[4+2\left(1-0\right)+\frac{\left(1-0\right)\left(2-0\right)}{6}\right]\\ =6+\frac{1}{3}=\frac{19}{3}\end{array}$

We know that  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}=\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[f\left(a\right)+f\left(a+h\right)+f\left(a+2h\right)+......+f\left(a+\left(n-1\right)h\right)\right]$

where nh = b - a

Here, a = 0, b = 5, nh = 5 and f(x) = x + 1

$\begin{array}{l}\therefore {\displaystyle \underset{0}{\overset{5}{\int }}\left(x+1\right)dx=\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[1+\left(h+1\right)+\left(2h+1\right)+.....+\left(\left(n-1\right)h+1\right)\right]}\\ \Rightarrow {\displaystyle \underset{0}{\overset{5}{\int }}\left(x+1\right)dx=\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[n+h\left(1+2+3.....+\left(n-1\right)\right)\right]}\\ \Rightarrow {\displaystyle \underset{0}{\overset{5}{\int }}\left(x+1\right)dx=\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[nh+h\frac{n\left(n-1\right)}{2}\right]}\\ =\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}\left[nh+\frac{nh\left(nh-h\right)}{2}\right]\end{array}$

$\begin{array}{l}=\underset{h\to 0}{lim}\left[5+\frac{5\left(5-h\right)}{2}\right]\\ =\left[5+\frac{5\left(5-0\right)}{2}\right]\\ =5+\frac{25}{2}=\frac{35}{2}\end{array}$

We know that  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}=\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[f\left(a\right)+f\left(a+h\right)+f\left(a+2h\right)+......+f\left(a+\left(n-1\right)h\right)\right]$

where  $nh=b-a$

Here, a = a, b= b and f(x) = x

$\therefore {\displaystyle \underset{a}{\overset{b}{\int }}xdx}=\underset{h\to 0}{lim}h\left[a\left(a+h\right)+\left(a+2h\right)+......+\left(a+\left(n-1\right)h\right)\right]$

$\Rightarrow {\displaystyle \underset{a}{\overset{b}{\int }}xdx}=\underset{h\to 0}{lim}h\left[na+\left(1+2+3+.....+\left(n-1\right)\right)\right]$

$\Rightarrow {\displaystyle \underset{a}{\overset{b}{\int }}xdx}=\underset{h\to 0}{lim}h\left[anh+h\frac{n\left(n-1\right)}{2}\right]$

$=\underset{h\to 0}{lim}h\left[anh+\frac{nh\left(nh-h\right)}{2}\right]$

$=\underset{h\to 0}{lim}h\left[a\left(b-a\right)+\frac{\left(b-a\right)\left(b-a-h\right)}{2}\right]\left[?nh=b-a\right]$

$\begin{array}{l}=\left[a\left(b-a\right)+\frac{\left(b-a\right)\left(b-a\right)}{2}\right]\\ =\left(b-a\right)\left[a+\frac{b-a}{2}\right]\\ =\left(b-a\right)\left[\frac{2a+b-a}{2}\right]\\ =\frac{\left(b-a\right)\left(b+a\right)}{2}\\ =\frac{b^2-a^2}{2}\end{array}$

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution