**– Locate roots of f(x)=0 by the change in sign**

If *y* = *f*(*x*) changes sign between *x* = a and *x* = b and *f* (*x*) is continuous in this region then a root of *f *(*x*) = 0 lies between *x* = a and *x* = b.

Example : The equation f(x) = x^{3} -8x – 10 has a root between x=3 and x=4 as f(3)=27-24-10=-7 and f(4)=64-32-10=22

f(x) changes sign and f(x) is continuous, therefore there is a root between 3 and 4.

**– Solve equations by simple iterative methods**

Many equations of the form f (x) = 0 can be re–arranged as x = g (x) which leads to the iteration x_{n+1} = g(x_{n})

If an equation is rearranged as *x* = g(*x*) and if there is a root *x* = *á *

then the iteration

(i) will converge if -1 < g ‘(α) < 1

(a) will converge without oscillating if 0 <g'(α)<1

(b) Will oscillate and converge if -1 < g'(α)<0

(ii) will diverge if g'(α)<-1 or if g'(α)>1

Example : The equation x^{3} -8x – 10 = 0 can be rearranged as

x = \(\sqrt{8x + 10}\)

Let x_{0} = 3 and use the iteration x_{n+1 = \(\sqr{8x + 10}\) to gi}

x_{1} = 3.2396

x_{2} = 3.23994………..

And so on

**– Newton- Rhapson ‘s Method**

The Newton-Rhapson’s method uses an iterative method to calculate the roots of the equation f(x)=0

Suppose the there is a root x_{0} , then

x_{1} = x_{0}= \(\frac{f(x)}{f'(x)}\)

And the process is repeated

\(x_n - \frac{f(x)}{f'(x)}\)**– Numerical Integration**

The Trapezium rule

The area under the curve y = f(x)is given by

A = \(\int f(x) \) ≈ \(\frac{1}{2}\)(y_{0}+ y_{n} + 2(y_{1}+ y_{2} + y_{3}+…………y_{n-1}))

**– Numerical Integration**

The Trapezium rule

The area under the curve y = f(x) is given by

\(\int_a^{f}f(x)dx\) ≈ \(\frac{1}{2}(y_0 + y_n + 2(y_1 + y_2 + y_3............y_{n-1))}\)

**– Numerical methods**

Evaluate \(\int^1_0{\sqrt{x + 2dx}}\) using the values at x=0, x=0.25, x=0.50, x=0.75, and x=1

Then calculate the areas of the trapezium for the overestimate and the underestimate.