Problem

(1 point) The following table gives the the approximate amount of emissions, $E$, of nitrogen oxides in millions of metric tons per year in the US. Let $t$ be the number of years since 1940 and $E=f(t)$.
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline$t$ & 1940 & 1950 & 1960 & 1970 & 1980 & 1990 \\
\hline$E$ & 6.6 & 9.8 & 13.2 & 18.2 & 21.3 & 19.5 \\
\hline
\end{tabular}
(a) Using left endpoints, estimate the integral:
\[
\int_{0}^{50} f(t) d t \approx
\]
(b) Using right endpoints, estimate the integral:
\[
\int_{0}^{50} f(t) d t \approx
\]

Be sure that you know what the units of your answer are, and what its meaning is!
(Original data from the Statistical Abstract of the US, 1992)

Answer

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Answer

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.

Steps

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.

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