Question

In trapezoidal rule ${\int }_a^{b}f\left(x\right)dx=\frac{h}{2}[y_0+2(y_1+y_2+.....y_{a-1})+y_a$. Write the value of $h$.

Answer 1

In trapezoidal rule area under a curve $f\left(x\right)$ is the interval $(a,b)$ i.e. ${\int }_a^{b}f\left(x\right)dx$ is given by

${\int }_a^{b}f\left(x\right)dx=\frac{h}{2}[y_0+2(y_1+y_2+....y_{n-1})+y_n]$.

Where $n$ is the number of sub-intervals in which we divide interval $(a,b)$.

and $h=\frac{b-a}{n}$

Hence, the value of $h$ is $\frac{b-a}{n}$