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?

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Answer 1 · 2015-09-26T00:25:06

Refer to 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 $f(0)=0$
$f(1/2)=5/8$
$f(1)=2$
$f(3/2)=39/8$
$f(2)=10$

So we have that

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

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