Question

Use the concept of the definite integral to find the total area between the graph off(x) and thex-axis, by taking the limit of the associated right Riemann sum. Write the exact answer. Do not round. (Hint: E)f(x) < 0. Remember that the definite integral represents a signed area.)

Answer 1

Given the definite integral to find the total area between the graph of f(x)

`f(x)=3x+3\text{ on }\lbrack0,4\rbrack`

We first want to set up a Riemann sum. Based on the limits of integration, we have a = 0 , b = 4

`Fori=0,1,2,\ldots,n,letP=\mleft\{\xi\mright\}bearegularpartitionof\mleft[0,4\mright].Then`

Since we are using a right-endpoint approximation to generate Riemann sums, for each i, we need to calculate the function value at the right endpoint of the interval [xi−1,xi]. The right endpoint of the interval is xi, and since P is a regular partition,

`\begin{gathered} \Delta x=(b-a)/(n) \\ \Delta x=(4-0)/(n)=(4)/(n) \\ x_i=x_0+i\Delta x=0+i(4)/(n) \end{gathered}`

Thus, the function value at the right endpoint of the interval is

`\begin{gathered} f(x_i)=3((i4)/(n))+3 \\ f(x_i)=(12i)/(n)+3 \end{gathered}`

Then the Riemann sum takes the form

`\sum ^n_(i=1)f(x_i)\Delta x=\sum ^n_(i=1)((12i)/(n)+3_{})\Delta x`

Therefore,

`\sum ^n_(i=1)f(x_i)\Delta x=\sum ^n_(i=1)(12i)/(n)_{}\Delta x+\sum ^n_(i=1)3_{}\Delta x`

Hence,

`\sum ^n_(i=1)f(x_i)\Delta x=(12\Delta x)/(n)\sum ^n_(i=1)i_{}+3_{}\Delta x\sum ^n_(i=1)1`

Thus,

`\begin{gathered} \sum ^n_(i=1)f(x_i)\Delta x=(12\Delta x)/(n)*(n(n+1))/(2))+3_{}\Delta x* n \\ \\ \sum ^n_(i=1)f(x_i)\Delta x=(24)/(n)*((n+1))/(1))+12 \\ \sum ^n_(i=1)f(x_i)\Delta x=(24(n+1))/(n))+12=24(1+(1)/(n))+12 \end{gathered}`

Hence,

`\lim _(n\to\infty)\sum ^n_(i=1)f(x_i)\Delta x=\lim _(n\to\infty)24(1+(1)/(n))+12=24+12=36`
`\begin{gathered} \text{ Since }\lim _(n\to\infty)\sum ^n_(i=1)f(x_i)\Delta x=\int ^4_03x+3dx\text{ then} \\ \int ^4_03x+3dx=36 \end{gathered}`

Therefore the definite integral for the function is 36