\underline{Quiz 2}

Davis Name: M211 Pledge:

(9pts.) 1. Use integration tables to integrate the following:

a. ∫ \frac{y^{2}+4y+6}{y^{2}+4y+5}dy

y + arctan (y + 2) + C

b. ∫ (x^{3} − 5x + 2)e^{−2x}dx

−e^{−2x}(\frac{1}{2}(x^{3} − 5x + 2) + \frac{1}{4}(3x^{2} − 5) + \frac{1}{8}(6x) + \frac{6}{16})

c. ∫ sin^{4} (x) cos^{2} (x)dx

\frac{1}{6} sin^{5} x cos x − \frac{1}{24} sin^{3} x cos x − \frac{1}{16} sin x cos x + \frac{1}{16}x + C

(11pts.) 2. Given the picture of y = e^{−.01x^{2}}, set up (but do not evaluate) sums which can be used to approximate the area under the curve from 1 to 3, subdividing the interval into 8 regions, using:

a. Left-hand Riemann sum /Library/classes/212f99/quiz2pic.eps /Library/classes/212f99/quiz2pic.eps

/Library/classes/212f99/quiz2pic.epsGraphic file (type eps)

∑_{i=0}^{7} f(1 + \frac{i}{4})\frac{1}{4}: this is an overestimate because the rectangles hang over the curve (draw picture).

b. Right-hand Riemann sum

∑_{i=1}^{8} f(1 + \frac{i}{4})\frac{1}{4}: this is an underestimate because the rectangles fall below the curve (draw picture).

c. Trapezoid rule

\frac{LHS+RHS}{2} : when you connect the endpoints, the trapezoid falls below the curve, so this is an underestimate.

d. Midpoint rule

∑_{i=0}^{7} f(1+ \frac{2i+1}{8})\frac{1}{4}: this is an overestimate because the trapezoid with top edge the tangent to the curve at the midpoint hangs over the curve (draw picture). e. Simpson’s rule

\frac{2Mid(8)+T rap(8)}{3}

In each case (except Simpson’s rule), indicate whether the estimate is an overestimate or an underestimate and explain why (based on the picture).