MATH 156 Lab 11Topic 1: Comparing integrals.We recall that if NiMyLSUiZkc2IyUieEctJSJnR0Ym for all NiMtJSNpbkc2JCUieEc3JCUiYUclImJH, then NiMtJSRpbnRHNiQtJSJmRzYjJSJ4Ry9GKTslImFHJSJiRw== < NiMtSSRJbnRHNiI2JC1JImdHRiU2I0kieEdGJS9GKjtJImFHRiVJImJHRiU=. This inequality allows us to estimate some integrals that are difficult to calculate.Show that NiMtSSRpbnRHNiI2JComSSJ4R0YlIiIiLUkjbG5HRiU2IywmRilGKS1JJXNxcnRHRiU2IywmRilGKSokRigiIiNGKUYpRikvRig7IiIhRik= < NiMtSSNsbkc2IjYjLCYiIiJGKC1JJXNxcnRHRiU2IyIiI0Yo.First we graph the function to integrate on the given interval.f:=x->x*ln(1+sqrt(1+x^2));

with(plots):plot(f(x), x=0..1);

From the graph it is obvious that it is an increasing function. Consequently, NiMyLSUiZkc2IyUieEctRiU2IyIiIg== for NiMyJSJ4RyIiIg==. Now we can take NiMvLSUiZ0c2IyUieEctJSNsbkc2IywmIiIiRiwtJSVzcXJ0RzYjIiIjRiw=, since NiMvLSUiZkc2IyIiIi0lI2xuRzYjLCZGJ0YnLSUlc3FydEc2IyIiI0Yn.The integral NiMtJSRpbnRHNiQtJSJnRzYjJSJ4Ry9GKTsiIiEiIiI= is equal to NiMtJSNsbkc2IywmIiIiRictJSVzcXJ0RzYjIiIjRic=, as the function is constant and the interval has length 1.

Show that NiMyLSUmRmxvYXRHNiQiJHYkISIkLSUjbG5HNiMtRiU2JCIjOiEiIg==. Here are the steps to follow: Recall that NiMvLSUjbG5HNiMtJSZGbG9hdEc2JCIjOiEiIi0lJGludEc2JComIiIiRjAlInhHRisvRjE7RjBGJw==. Plot on the same graph the functions NiMvLSUiZ0c2IyUieEcqJiIiIkYpRichIiI= and NiMvLSUiZkc2IyUieEcsJiIiIyIiIkYnISIi. What do you notice? Explain why NiMvLSUkaW50RzYkLCYiIiMiIiIlInhHISIiL0YqO0YpLSUmRmxvYXRHNiQiIzpGKy1GLzYkIiR2JCEiJA==.

Topic 2: Improper integrals.We plot first on the same graph the functions NiMvLSUiZkc2IyUieEcqJiIiIkYpKiRGJyIiIyEiIg== and NiMvLSUiZ0c2IyUieEcqJiIiIkYpRichIiI=. Just by looking at the graphs over a long interval, we cannot decide which improper integralNiMtJSRpbnRHNiQqJiIiIkYnJSJ4RyEiIi9GKDtGJyUpaW5maW5pdHlH or NiMtJSRpbnRHNiQqJiIiIkYnKiQlInhHIiIjISIiL0YpO0YnJSlpbmZpbml0eUc= converges or diverges. However, it becomes clear that NiMyKiYiIiJGJSokJSJ4RyIiIyEiIiomRiVGJUYnRik= for all NiMyIiIiJSJ4Rw== and, consequently, the area below the graph of NiMqJiIiIkYkJSJ4RyEiIg== is larger than the area below the graph of NiMqJiIiIkYkKiQlInhHIiIjISIi.reset:f:=x->1/x^2; g:=x->1/x;

plot([f(x),g(x)], x=1..100, color=[blue, red]);

Putting the names of the functions in square brackets assigns this order to the plots. The color command assigns the color blue to the graph NiMvLSUiZkc2IyUieEcqJiIiIkYpKiRGJyIiIyEiIg== and the color red to the graph NiMvLSUiZkc2IyUieEcqJiIiIkYpKiRGJyIiIyEiIg==. With Maple we can compute the improper integrals and see that NiMtJSRpbnRHNiQqJiIiIkYnJSJ4RyEiIi9GKDtGJyUpaW5maW5pdHlH=NiMlKWluZmluaXR5Rw==, while NiMtJSRpbnRHNiQqJiIiIkYnKiQlInhHIiIjISIiL0YpO0YnJSlpbmZpbml0eUc==1.A:=Int(1/x,x = 1 .. infinity);

value(A);

B:=Int(1/x^2,x = 1 .. infinity);

value(B);

To understand the convergence or divergence of these improper integrals, we compute values of the integrals NiMtJSRpbnRHNiQqJiIiIkYnJSJ4RyEiIi9GKDtGJyUiYkc= and NiMtJSRpbnRHNiQqJiIiIkYnKiQlInhHIiIjISIiL0YpO0YnJSJiRw== for various NiMlImJH. We can do this effectively with a loop.for j from 1 to 20 do b:=10^j: evalf(int(g(x), x=1..b)), evalf(int(f(x), x=1..b)): od;

Make a table of values for the improper integrals NiMtJSRpbnRHNiQtJSRleHBHNiMsJComLSUmRmxvYXRHNiQiIiYhIiIiIiIlInhHRjBGLy9GMTsiIiMlKWluZmluaXR5Rw== and NiMtJSRpbnRHNiQtJSRleHBHNiMsJComLSUmRmxvYXRHNiQiIiIhIiJGLiUieEdGLkYvL0YwOyIiIyUpaW5maW5pdHlH. Can you decide whether they converge or diverge?

They both seem to converge. In fact it seems that NiMtJSRpbnRHNiQtJSRleHBHNiMsJComLSUmRmxvYXRHNiQiIiYhIiIiIiIlInhHRjBGLy9GMTsiIiMlKWluZmluaXR5Rw===0.7357588823 and NiMtJSRpbnRHNiQtJSRleHBHNiMsJComLSUmRmxvYXRHNiQiIiIhIiJGLiUieEdGLkYvL0YwOyIiIyUpaW5maW5pdHlH=8.187307531. Evaluate the integrals by hand and show that these answers are true.

Topic 3: Comparing improper integrals.Plot on the same graph NiMvLSUiZkc2IyUieEcqJiIiIkYpKUYnLSUmRmxvYXRHNiQiI0QhIiJGLw== and NiMvLSUiZ0c2IyUieEcqJiIiIkYpKiRGJyIiIyEiIg==. Explain why the improper integral NiMtJSRpbnRHNiQqJiIiIkYnKSUieEctJSZGbG9hdEc2JCIjRCEiIkYuL0YpO0YnJSlpbmZpbml0eUc= converges.

.Decide whether the improper integral NiMtJSRpbnRHNiQqJiwmLSUkc2luRzYjJSJ4RyIiIiIiJEYsRiwtJSVzcXJ0R0YqISIiL0YrO0YsJSlpbmZpbml0eUc= converges or not.

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