I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
Section 5.5 : Area Problem
For problems 1 – 5 estimate the area of the region between the function and the x-axis on the given interval using \(n = 6\) and using,
- the right end points of the subintervals for the height of the rectangles,
- the left end points of the subintervals for the height of the rectangles and,
- the midpoints of the subintervals for the height of the rectangles.
- \(f\left( x \right) = 15 + 4x - {x^3}\) on \(\left[ {1,3} \right]\)
- \(g\left( x \right) = - 3{x^2} + 2x - 1\) on \(\left[ { - 4,0} \right]\)
- \(h\left( x \right) = 8\ln \left( x \right) - x\) on \(\left[ {2,6} \right]\)
- \(f\left( x \right) = {\sin ^2}\left( {\frac{x}{2}} \right)\) on \(\left[ {0,3} \right]\)
- \(g\left( x \right) = \sin \left( x \right)\cos \left( x \right) - 1\) on \(\left[ { - 2,1} \right]\)
For problems 6 – 8 estimate the net area between the function and the x-axis on the given interval using \(n = 8\) and the midpoints of the subintervals for the height of the rectangles. Without looking at a graph of the function on the interval does it appear that more of the area is above or below the x-axis?
- \(h\left( x \right) = 8x - \sqrt {x + 4} \) on \(\left[ { - 3,2} \right]\)
- \(g\left( x \right) = 5 + x - {x^2}\) on \(\left[ {0,4} \right]\)
- \(f\left( x \right) = x{{\bf{e}}^{ - {x^{\,2}}}}\) on \(\left[ { - 1,1} \right]\)