Step 1 :Using left endpoints, we estimate the integral by approximating the area under the curve as a series of rectangles. The width of each rectangle is the time interval (10 years), and the height is the emission at the start of each interval.
Step 2 :The integral from 0 to 50 is approximately: \[\int_{0}^{50} f(t) d t \approx (10)(6.6) + (10)(9.8) + (10)(13.2) + (10)(18.2) + (10)(21.3)\]
Step 3 :Calculating the above expression: \[\int_{0}^{50} f(t) d t \approx 660 + 980 + 1320 + 1820 + 2130 = 6910\]
Step 4 :\(\boxed{\text{So, the approximate integral from 0 to 50 using left endpoints is 6910 million metric tons.}}\)
Step 5 :Using right endpoints, we estimate the integral by approximating the area under the curve as a series of rectangles. The width of each rectangle is the time interval (10 years), and the height is the emission at the end of each interval.
Step 6 :The integral from 0 to 50 is approximately: \[\int_{0}^{50} f(t) d t \approx (10)(9.8) + (10)(13.2) + (10)(18.2) + (10)(21.3) + (10)(19.5)\]
Step 7 :Calculating the above expression: \[\int_{0}^{50} f(t) d t \approx 980 + 1320 + 1820 + 2130 + 1950 = 7200\]
Step 8 :\(\boxed{\text{So, the approximate integral from 0 to 50 using right endpoints is 7200 million metric tons.}}\)
Step 9 :The units of these answers are million metric tons, and they represent the total amount of nitrogen oxides emitted in the US from 1940 to 1990.