14.9. Area Under a Curve http://www.ck12.org
- While a graph is helpful to visualize the problem and drawing each box can help give meaning to each summand,
it is not always necessary. Since there are going to be 8 subintervals over the total interval of− 1 ≤x≤7, each
interval is going to have a width of 1. The height of each interval is going to be at the right hand endpoints of each
subinterval (0, 1, 2, 3, 4, 5, 6, 7).
∑height·width=
7
i∑= 0 (^3 i^2 −^1 )·^1 =^412- Each interval will be only^12 wide which means that the left endpoints havexvalues of: 2, 2.5, 3, 3.5, 4, 4.5, 5, 5.5.
Since the index of summation notation does not work with decimals, double each of these to get a good counting
sequence: 4, 5, 6, 7, 8, 9, 10, 11 and remember to halve them in the argument of the summation.
x= 2 i∑height·width=
11
∑i= 4(
(^4
2 i))
·^12 ≈ 4. 7462
- When the number of subintervals gets large and the subintervals get extremely narrow it will be impossible to
draw an accurate picture. This is why using summation notation and thinking through what the indices and the
argument will be is incredibly important. With 20 subintervals between[ 1 , 3 ], each interval will be 0.1 wide. Left
endpoints means that the first box has a height off( 1 )and the second box has a height off( 1. 1 ).
∑height·width=f(^1 )·^0.^1 +f(^1.^1 )·^0.^1 +f(^1.^2 )·^0.^1 +···+f(^2.^9 )·^0.^1
= 0. 1 (f( 1 )+f( 1. 1 )+···f( 2. 9 ))
= 0. 1 ·29
i∑= 10 f(i
10)
= 0. 1 ·
29
i∑= 10(i
10)( 10 i)≈ 12. 47144
Your calculator can compute summations when you go under the math menu.
Practice
- Approximate the area under the curve using eight subintervals and right endpoints.