# How do you use the trapezoidal rule with n=4 to approximate the area between the curve x ln(x+1) from 0 to 2?

Aug 13, 2018

$\frac{1}{2} \ln 6 \approx 0.9$

#### Explanation:

The trapezoidal rule states that the area under an integral can be approximated by the equation:

${\int}_{a}^{b} f \left(x\right) \setminus \mathrm{dx} \approx \frac{\Delta x}{2} \left[f \left({x}_{0}\right) + 2 f \left({x}_{1}\right) + 2 f \left({x}_{2}\right) + 2 f \left({x}_{3}\right) + \ldots + 2 f \left({x}_{n - 1}\right) + f \left({x}_{n}\right)\right]$

where:

• $\Delta x = \frac{b - a}{n}$

• $n$ is the number of trapezoids

• ${x}_{0} = a$

• ${x}_{1} , {x}_{2} , \ldots , {x}_{n}$ are equally spaced $x$-coordinates of the right edges of trapezoids $1 , 2 , 3 , \ldots , n$.

So, we get:

${\int}_{0}^{2} x \ln \left(x + 1\right) \setminus \mathrm{dx} \approx \frac{b - a}{2 n} \left[f \left(0\right) + 2 f \left(1\right) + f \left(2\right)\right]$

$= \frac{2 - 0}{2 \cdot 4} \left[f \left(0\right) + 2 f \left(1\right) + f \left(2\right)\right]$

$= \frac{2}{8} \left[0 \ln 1 + 2 \left(1 \ln 2\right) + 2 \ln 3\right]$

$= \frac{1}{4} \left(2 \ln 2 + 2 \ln 3\right)$

$= \frac{1}{2} \ln 6$

$\approx 0.9$