Calculus II Class 02/03


Calculus I Quiz 0203

Find the following: 

antiderivative of f(t) = 1 / t^2 is F(t) = -1/t.

Integral from .1 to 1 is therefore F(1) - F(.1) = -1/1 - (-1/.1) = -1 + 10= 9.

This represents the area under the y = 1/t^2 curve, from t = .1 to t = 1.  The corresponding segment of the curve is concave up and decreases from the point (.1, 100) to the point (1, 1).

 

Integral from .1 to 1 is therefore F(1) - F(.01) = -1/1 - (-1/.01) = -1 + 100 = 99.

This represents the area under the y = 1/t^2 curve, from t = .01 to t = 1.  The corresponding segment of the curve is concave up and decreases from the point (.01, 10,000) to the point (1, 1).

Integral from .001 to 1 is therefore F(1) - F(.001) = -1/1 - (-1/.001) = -1 + 1000 = 999.

This represents the area under the y = 1/t^2 curve, from t = .001 to t = 1.  The corresponding segment of the curve is concave up and decreases from the point (.001, 1,000,000) to the point (1, 1).

Integral from x to 1 is F(1) - F(x) = -1/1 - (-1/x) = -1 + 1/x = 1/x - 1 = (1-x) / x.

This represents the area under the y = 1/t^2 curve, from t = x to t = 1.  The corresponding segment of the curve is concave up and decreases from the point (x, 1/x^2) to the point (1, 1).

The integral is (1-x) / x, which as x -> 0 approaches form 1 / 0, indicating that the integral approaches infinity.

This means that if we let x be close enough to zero, the integral can exceed any bound we wish to specify.  We can get as much area as we like by letting x be sufficiently close to zero.

 

Find the following: 

You can't find an antiderivative in closed for; no such closed form exists for this funciton.

This function goes through a complete cycle every time 2 pi / t changes by 2 pi.

The function will go through increasing numbers of cycles every time we move twice as close to t = 0.

Without an antiderivative function it will be difficult to even approximate the function.  However we can see that as we move toward t = 0 we will be accumulating first positive, then negative, then positive, then negative 'areas'.  We also note that the 'areas' are increasingly 'bunched up'.  

 

Find the following: 

As seen in the figure below, substitution u = 2 pi / t leads us to the antiderivative 

and then to 

the integral 

 

this time we get the integral 

 

.02986

Does the result keep fluctuating or does it start behaving in such a way that we do indeed get a limit?

Numerical investigation is unclear.

Evaluating the integral from x to 1, where r is a number randomly chosen between 0 and .001, we obtain the 10 independent approximations 

Evaluating the integral from x to 1, where r is a number randomly chosen between 0 and .0001, we obtain the 10 independent approximations 

Evaluating the integral from x to 1, where r is a number randomly chosen between 0 and .00001, we obtain the 10 independent approximations 

 

Evaluating the integral from x to 1, where r is a number randomly chosen between 0 and .000001, we obtain the 10 independent approximations 

Evaluating the integral from x to 1, where r is a number randomly chosen between 0 and .001, we obtain the 10 independent approximations 

It is notable that none of the approximations is negative and that there is no consistency to the results.  It is conceivable that no matter how close we place r to zero, the value of the integral will be unpredictable.

It would then follow that as we allow x to approach zero, the integral of this function from x to 1 will still tend to fluctuate pretty randomly between 0 and at least .3.

Note that these are 10-significant-figure approximations made by DERIVE.  It is conceivable that the fluctuations observed are the result of the nature of the approximation scheme used by that program rather than the actual behavior of the functoin.

 

Find the following: 

The antiderivative F(t) = - 1 / t gives us 

 

The antiderivative F(t) = - 1 / t gives us 

 

The antiderivative F(t) = - 1 / t gives us 

Bigger and bigger 'jumps' in the upper limit result in smaller and smaller change in the integral.  Looks like as the upper limit approaches infinity the integral will converge to 1.

The antiderivative F(t) = - 1 / t gives us 

As x gets larger and larger, -1/x approaches zero and our integral approaches 0 + 1 = 1.

 

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Sketch the graph of y = 1 / (t - 2)^2 from t = 0 to t = 4.