Question

The function $f\left(x\right)=2x+3$is defined on the interval role="math" localid="1652696446205" $\left[0,4\right]$.

(a) Graph f.

In (b)–(e), approximate the area A under f from x = 0 to x = 4 as follows:

(b) Partition $\left[0,4\right]$into four subintervals of equal length and choose u as the left endpoint of each subinterval.

(c) Partition $\left[0,4\right]$into four subintervals of equal length and choose u as the right endpoint of each subinterval.

(d) Partition $\left[0,4\right]$into eight subintervals of equal length and choose u as the left endpoint of each subinterval.

(e) Partition $\left[0,4\right]$into eight subintervals of equal length and choose u as the right endpoint of each subinterval.

(f) What is the actual area A?

Short answer

Part (a)

Part (b)

Part (c)

Part (d)

Part (e)

Part (f) The area is 28 square units.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given function.

$f\left(x\right)=2x+3$

Part (b) Step 2. Draw the graph.

Plot the given function at the given interval.

Part (b) Step 1. Draw the graph.

Draw the graph of the given function with four subintervals and choose uas the left endpoint

Part (c) Step 1. Draw the graph.

Draw the graph of the given function with four subintervals and choose u as the right endpoints.

Part (d) Step 1. Draw the graph.

Draw the graph of the given function with eight subintervals and choose u as the left endpoints.

Part (e) Step 1. Draw the graph.

Draw the graph of the given function with eight subintervals and choose u as the right endpoints.

Part (f) Step 1. Find the area.

Calculate the area for the given interval.

$\begin{array}{rcl}A& =& {\int }_a^{b}f\left(x\right)dx\\ & =& {\int }_0^4\left(2x+3\right)dx\\ & =& {\left[x^2+3x\right]}_0^4\\ & =& 16+12\\ & =& 28\end{array}$