Split up the interval [0, 3] into 3 equally spaced subintervals of length . So we have the partition
[0, 1] U [1, 2] U [2, 3]
The left endpoint of the -th subinterval is
where .
Then the area is given by the definite integral and approximated by the left-hand Riemann sum
x + 2 ≥ 6
<h3>Find</h3>the graph for the solution
<h3>Solution</h3>We can solve for x by subtacting 2 from both sides of the inequality.
... x + 2 -2 ≥ 6 -2
... x ≥ 4 . . . . . . . . . . simplify
The answer is read, "x is greater than or equal to 4." This means the solution is those real numbers that are to the right of 4 on the number line, and including 4 on the number line. Numbers to the left of 4 are <em>not included</em>. The inclusion of the number 4 is shown by putting a solid dot there (not an open circle).
The appropriate choice is the 3rd one.
