You would have four trapezoids in your approximation.
the width of each of those traps is .5
and the heights at
x = 1, 1.5, 2, 2.5, and 3 are
h = 8, 6.75, 5, 2.75, and 0 respectively
the area is appr. = .5/2[(8+6.75)+(6.75+5)+(5+2.75)+(2.75+0)]
= 9.25
(the exact answer would have been 9.33333..)
do the second one the same way.
Approximate the area under each curve by evaluating te function at the left-hand endpoints of the subintervals.
1. f(x)=9-x^2 from x=1 to x=3; 4 subintervals.
2. f(x)=x^2 +x+ from x= -1 to x=1; 4 subintervals.
1 answer