(a) Graph the functions f(x)=1 / x . and g(x)=1 / x^0.9 in...

(a) Graph the functions $f(x)=1 / x$ . and $g(x)=1 / x^{0.9}$ in the viewing rectangles $[0,10]$ by $[0,1]$ and $[0,100]$ by $[0,1] .$ (b) Find the areas under the graphs of $f$ and $g$ from $x=1$ to $x=t$ and evaluate for $t=10,100,10^{4}, 10^{6}, 10^{10},$ and $10^{20} .$ (c) Find the total area under each curve for $x \geqslant 1,$ if it exists.

(a) Graph the functions $f(x)=1 / x$ . and $g(x)=1 / x^{0.9}$ in the viewing rectangles $[0,10]$ by $[0,1]$ and $[0,100]$ by $[0,1] .$
(b) Find the areas under the graphs of $f$ and $g$ from $x=1$ to $x=t$ and evaluate for $t=10,100,10^{4}, 10^{6}, 10^{10},$ and $10^{20} .$
(c) Find the total area under each curve for $x \geqslant 1,$ if it exists.

Key Concepts

Parameterization in Integration

Parameterization in integration involves using variables, such as an upper limit t, to express how accumulated area changes as the limit extends. This technique helps in analyzing and comparing the growth of the area under different curves over increasing intervals, highlighting differences in convergence behavior.

p-Integrals and Convergence Criteria

p-Integrals are of the form ?1/x? dx and play a central role in determining convergence, based on the exponent p. They converge to a finite value if p > 1 and diverge if p ? 1, a principle which is crucial for understanding the behavior of areas under curves of reciprocal functions and similar forms.

Reciprocal Functions

Reciprocal functions, which take the form 1/x?, have characteristic traits such as vertical asymptotes at x = 0 and horizontal asymptotes as x tends to infinity. These properties help in predicting the function’s behavior over different domains and are foundational in the study of rational functions.

Function Graphing

Graphing functions involves visualizing their behavior within specified coordinate ranges, identifying key features such as intercepts and asymptotes. This is essential for understanding how functions like 1/x change over given intervals and for appreciating differences in shape and scale between similar functions.

Improper Integrals

Improper integrals are used to determine areas under curves extending over unbounded intervals, typically handled by taking limits. This concept is critical for evaluating the total area under functions such as 1/x? from a finite starting point to infinity, where convergence or divergence must be assessed.

Transcript

00:01 In this problem, we're given the functions f of x equals 1 over x to the 1 .1, and g of x equals 1 over x to 0 .9.

00:10 And we're asked a few questions about these functions.

00:13 The first thing they ask us to do is to graph both of these with different viewing rectangles.

00:20 So in my ti -84 plus, i've already graphed these, and you can see the graphs here.

00:26 The graph on your left is with the y range from zero to 10 and the one on the right is the y range from zero to 100 and the way i changed that is i changed the y max and y min depending on the viewing rectangle that the problem wanted and if we scroll down here's those graphs for the g of x equation again the one the left is that the viewing rectangle from 0 to 10 and the one on the right is the viewing rectangle from 0 to 100.

01:04 In part b, they ask us to find the area under these graphs.

01:10 So do that, we're going to go ahead and take the integral.

01:13 I'm going to start with the f of x equation.

01:18 So that one.

01:20 And to do that, we're just going to take the integral.

01:22 I'm going to call it a sub t because they want us to take this from x equals 1 to x equals t.

01:31 So if we do that, we get the integral from t to 1 of 1 over x to the 1 .1 d x.

01:43 And from here, if we take this integral, we get negative 1 over 0 .1 x to the 0 .1.

01:58 And we are going to evaluate this from 1 to t.

02:03 So if we plug in these values, we get 10 minus 1 over 0 .1 t to the 0 .1.

02:19 And from here, they ask us to evaluate with t equaling a lot of different values.

02:25 So if we have t equal 10, and you just plug that in to our expression above, for a.

02:38 10 or a of 10, you'll get 2 .057.

02:44 A, or with t equaling 100, we'll get 3 .690.

02:55 If we set a, i'm sorry, t equal to 10 to the fourth, we get 6 .019.

03:08 If we set t equal to 10 to the sixth, we'll get 7 .488.

03:26 A with t equaling 10 to the 10th will give us 9 exactly.

03:36 And finally with t equaling 10 to the 20th we get 9 .9...

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