# How do you use the trapezoidal rule with n=6 to approximate the area between the curve 9 sqrt (ln x)  from 1 to 4?

Oct 26, 2015

#### Answer:

See the explanation section below.

#### Explanation:

To approximate the Integral ${\int}_{a}^{b} f \left(x\right) \mathrm{dx}$ using trapezoidal approximation with $n$ intervals, use
${T}_{n} = \frac{\Delta x}{2} \left[f \left({x}_{0}\right) + 2 f \left({x}_{1}\right) + 2 f \left({x}_{2}\right) + \cdot \cdot \cdot 2 f \left({x}_{n - 1}\right) + f \left({x}_{n}\right)\right]$

In this question we have:
$f \left(x\right) = 9 \sqrt{\ln x}$
$\left\{a , b\right] = \left[1 , 4\right]$, and
$n = 6$.

So we get
$\Delta x = \frac{b - a}{n} = \frac{4 - 1}{6} = \frac{1}{2} = 0.5$

The endpoints of the subintervals are found by beginning at $a = 1$ and successively adding $\Delta x = 0.5$ to find the points until we get to ${x}_{n} = b = 4$.

${x}_{0} = 1$, ${x}_{1} = 1.5$, ${x}_{2} = 2$, ${x}_{3} = 2.5$, ${x}_{4} = 3$, ${x}_{5} = 3.5$, ${x}_{6} = 4 = b$,

Now apply the formula (do the arithmetic) for $f \left(x\right) = 9 \sqrt{\ln x}$.

${T}_{6} = \frac{\Delta x}{2} \left[f \left({x}_{0}\right) + 2 f \left({x}_{1}\right) + 2 f \left({x}_{2}\right) + \cdot \cdot \cdot 2 f \left({x}_{5}\right) + f \left({x}_{6}\right)\right]$