Question

Using a unit step size, the value of integral ${\int }_1^{2}xlnx$ dx by trapezoidal rule is

• A. 0.693

Answer 1

The correct option is A 0.693
Let y(x) = x lnx
Step size is 1
So, h = 1
$y_0=y\left(1\right)$ = ln1 = 0
$y_1=y\left(2\right)$ = 2ln2
Trapezoidal rule
${\int }_a^{b}y\left(x\right)dx=\frac{h}{2}[y_0+2(y_1+y_2+y_3+......y_{n-1})+y_n]$
here n = 1
so ${\int }_1^2$x lnx dx $=\frac{1}{2}[y_0+y_1]$
$=\frac{1}{2}$ [0+2 ln2]
= 0.693