# How do you use the trapezoidal rule with n=4 to approximate the area between the curve y(t)=(t^3 +t) from 0 to 2?

Refer to explanation

#### Explanation:

Hence we have n=4 we got 4 trapezoid hence we have that

Δx=(b-a)/n=(2-0)/4=1/2

From the trapezoidal rule we have that

int_0^2 (t^3+t)dt=(f(0)+f(1/2))/2*Δx+(f(1/2)+f(1))/2*Δx+(f(1)+f(3/2))/2*Δx+(f(3/2)+f(2))/2*Δx=(Δx)/2*[f(0)+2f(1/2)+2f(1)+2f(3/2)+f(2)]

Hence $f \left(0\right) = 0$
$f \left(\frac{1}{2}\right) = \frac{5}{8}$
$f \left(1\right) = 2$
$f \left(\frac{3}{2}\right) = \frac{39}{8}$
$f \left(2\right) = 10$

So we have that

int_0^2 (t^3+t)dt=(Δx)/2*[f(0)+2f(1/2)+2f(1)+2f(3/2)+f(2)]=(1/2)/2*[0+2*5/8+2*2+2*39/8+10]=1/4*(25)=6.25