Question

the 2000 census in a particular area gives us an age distribution that is approximately given (in millions) by the function
f(x)=39.3 + 2.09x - 0.708x²
where x varies from 0 to 9 decades. the population of a given age group can be found by integrating this function over the interval for that age group.
a. find the integral over the interval 0,9.
(round to the nearest integer as needed.)

Explanation:

Step1: Use the integral formula

$$\int_{a}^{b}(f(x)+g(x)-h(x))dx=\int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx-\int_{a}^{b}h(x)dx$$
For \(y = 39.3+2.09x - 0.708x^{2}\), \(\int_{0}^{9}(39.3 + 2.09x-0.708x^{2})dx=\int_{0}^{9}39.3dx+\int_{0}^{9}2.09xdx-\int_{0}^{9}0.708x^{2}dx\)

Step2: Integrate each term

• Integrate \(\int_{0}^{9}39.3dx\):

Using \(\int kdx=kx + C\) (\(k\) is a constant), \(\int_{0}^{9}39.3dx=39.3x\big|_{0}^{9}=39.3\times(9 - 0)=353.7\)

• Integrate \(\int_{0}^{9}2.09xdx\):

Using \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), \(\int_{0}^{9}2.09xdx=2.09\times\frac{x^{2}}{2}\big|_{0}^{9}=2.09\times\frac{9^{2}-0^{2}}{2}=2.09\times\frac{81}{2}=84.645\)

• Integrate \(\int_{0}^{9}0.708x^{2}dx\):

Using \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), \(\int_{0}^{9}0.708x^{2}dx=0.708\times\frac{x^{3}}{3}\big|_{0}^{9}=0.708\times\frac{9^{3}-0^{3}}{3}=0.708\times\frac{729}{3}=170.652\)

Step3: Calculate the result

\(\int_{0}^{9}(39.3 + 2.09x-0.708x^{2})dx=353.7+84.645 - 170.652=267.693\)

Answer:

\(268\)