$\text{The exact solution of}{\int }_0^4\frac{d}{x}\text{is represented as}\mathrm{ n.}$

$\text{If}\mathrm{ m}\text{ represents numerically evaluated value using trapezoidal rule with 4 equal subintervals in the range of x, then find (m − n).}$

The correct option is 0.074.

To find the value of $m-n$, we need to calculate the exact solution $n$ and the numerically approximated value $m$ using the trapezoidal rule.

Step 1: Calculate the exact solution ($n$)

The exact integral is given by:

$n={\int }_0^4\frac{dx}{1+x}$

Integrating the function, we get:

$n={\left[\ln \right(1+x\left)\right]}_0^4$

Substitute the limits of integration:

$n=\ln (1+4)-\ln (1+0)$

$n=\ln \left(5\right)-\ln \left(1\right)$

Since $\ln \left(1\right)=0$, we have:

$n=\ln \left(5\right)\approx 1.6094$

Step 2: Calculate the numerical value ($m$) using the trapezoidal rule

We are given that the interval is from $x=0$ to $x=4$ with $4$ equal subintervals ($N=4$).

The width of each subinterval ($h$) is calculated as:

$h=\frac{4-0}{4}=1$

The grid points are:
$x_0=0$
$x_1=1$
$x_2=2$
$x_3=3$
$x_4=4$

Now, we evaluate the function $f\left(x\right)=\frac{1}{1+x}$ at each grid point:
$y_0=f\left(0\right)=\frac{1}{1+0}=1$
$y_1=f\left(1\right)=\frac{1}{1+1}=0.5$
$y_2=f\left(2\right)=\frac{1}{1+2}\approx 0.3333$
$y_3=f\left(3\right)=\frac{1}{1+3}=0.25$
$y_4=f\left(4\right)=\frac{1}{1+4}=0.2$

Applying the trapezoidal rule formula:

$m=\frac{h}{2}[y_0+y_4+2(y_1+y_2+y_3\left)\right]$

Substitute the calculated values into the formula:

$m=\frac{1}{2}[1+0.2+2(0.5+0.3333+0.25\left)\right]$

$m=\frac{1}{2}[1.2+2(1.0833\left)\right]$

$m=\frac{1}{2}[1.2+2.1666]$

$m=\frac{1}{2}\left[3.3666\right]\approx 1.6833$

Step 3: Calculate the difference ($m-n$)

Subtracting the exact value from the numerical approximation:

$m-n=1.6833-1.6094=0.0739$

Rounding this to three decimal places yields:

$m-n\approx 0.074$