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?

Question

Samantha Adkison · Accepted Answer

#1/2ln6~~0.9#

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

#int_a^bf(x) \ dx~~(Deltax)/2[f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+...+2f(x_(n-1))+f(x_n)]#

where:

#Deltax=(b-a)/n#

#n# is the number of trapezoids

#x_0=a#

#x_1,x_2,...,x_n# are equally spaced #x#-coordinates of the right edges of trapezoids #1,2,3,...,n#.

So, we get:

#int_0^2xln(x+1) \ dx~~(b-a)/(2n)[f(0)+2f(1)+f(2)]#

#=(2-0)/(2*4)[f(0)+2f(1)+f(2)]#

#=2/8[0ln1+2(1ln2)+2ln3]#

#=1/4(2ln2+2ln3)#

#=1/2ln6#

#~~0.9#

Adam Adkins · Answer

undefined To use the trapezoidal rule with $ n = 4 $ to approximate the area between the curve $ x \ln(x + 1) $ from 0 to 2, follow these steps:

1. Divide the interval [0, 2] into 4 equal subintervals. Since $ n = 4 $, each subinterval will have a width of $ \Delta x = \frac{2 - 0}{4} = 0.5 $.
2. Calculate the function values at the endpoints and midpoint of each subinterval: $ x_0 = 0 $, $ x_1 = 0.5 $, $ x_2 = 1 $, $ x_3 = 1.5 $, $ x_4 = 2 $.
3. Evaluate $ f(x) = x \ln(x + 1) $ at each of these points to get the corresponding function values: $ f(x_0) $, $ f(x_1) $, $ f(x_2) $, $ f(x_3) $, $ f(x_4) $.
4. Use the trapezoidal rule formula to find the area under the curve within each subinterval:
   $$
   A_i = \frac{h}{2} \left( f(x_{i-1}) + f(x_i) \right)
   $$
   where $ h $ is the width of the subinterval and $ A_i $ is the area of the trapezoid for the $ i $-th subinterval.
5. Sum up all the individual areas $ A_i $ to get the total approximate area under the curve.

Using the given formula and the calculated function values at each endpoint and midpoint of the subintervals, you can find the approximate area under the curve between 0 and 2 using the trapezoidal rule with $ n = 4 $.