MAT136H1 Chapter Notes - Chapter 3b: Partial Fraction Decomposition, Antiderivative

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25 Jul 2018

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MATH136 – Lecture 16 – Integration by Partial Fraction

Mathematics: Reducing Problems

Example: Reducing partial problem to find an integral:

Once we have an approximation for finding a quantity on a slice, we can use out “slicing”

process to obtain an integral.

Example: Substitution and Integration by Parts:

Reduce a complex integral computation to a simpler one that we already know how to solve

with “guess and check”.

Today: we know how to integrate fractions like 

 via substitution and 

. How can we

reduce the problem of computing integral of rational function to this one?

Strategies for Integrating Rational Functions 



• If the degree of  degree of , try long division and use the method of partial

fractions on the remainder.

• If  is the product of distinct linear factors, use partial fractions of the form 



• If  contains repeated linear factor   , use partial fractions of the form



   

  

• If  contains an unfactorble quadratic , try a partial fraction of the form 



Question: Where a trigonometric function?

For which of the following functions, do you expect to see a trigonometric (pr inverse

trigonometric) function its anti-derivative

A. 



B. 



C. 



D. 



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MAT136H1 Full Course Notes

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Math136 lecture 16 integration by partial fraction. Example: reducing partial problem to find an integral: O(cid:374)ce (cid:449)e ha(cid:448)e a(cid:374) app(cid:396)o(cid:454)i(cid:373)atio(cid:374) fo(cid:396) fi(cid:374)di(cid:374)g a (cid:395)ua(cid:374)tit(cid:455) o(cid:374) a slice, (cid:449)e ca(cid:374) use out (cid:862)slici(cid:374)g(cid:863) process to obtain an integral. Today: we know how to integrate fractions like (cid:2869)(cid:3028)+(cid:3029) via substitution and (cid:2869)(cid:4666)(cid:4667). If the degree of (cid:1842)(cid:4666)(cid:4667) degree of (cid:1843)(cid:4666)(cid:4667), try long division and use the method of partial. If (cid:1843)(cid:4666)(cid:4667) is the product of distinct linear factors, use partial fractions of the form (cid:3002) (cid:3004) If (cid:1843)(cid:4666)(cid:4667) contains repeated linear factor (cid:4666) (cid:1855)(cid:4667), use partial fractions of the form. If (cid:1843)(cid:4666)(cid:4667) contains an unfactorble quadratic (cid:4666)(cid:4667), try a partial fraction of the form (cid:3002)+(cid:3003)(cid:4666)(cid:4667) (cid:1827)(cid:2869) (cid:1855)+ (cid:1827)(cid:2870) (cid:4666) (cid:1855)(cid:4667)(cid:2870) fractions on the remainder. Answer is d, because the term (cid:4666)(cid:2870)++(cid:886)(cid:4667) in denominator can not be factored/simplified so that you would get an integral for the form (cid:2869)(cid:3118)+(cid:3028)(cid:3118)(cid:1856) as one of the integrals when the integral after completing the square is split.

Related Questions

Page 3 of 5 which gives us y as a function of Theoretically. the integration on the right of t= 1/ - g + alpha f(V)dv can be performed and the result could provide velocity. v,. as a function of t; Then we could perform the integration in y=y0 + V/- g + alpha f(V)dv and substitute v(t) to obtain y as a function of t. Theoretically, that is. Let's look at in example An Example Suppose the air resistance model for a particular object in free fall has where units for v area feet/second Note how f is defined in terms of the sign of v. As usual. downward has been selected as the negative direction so velocities v of an object in free fall will have negative values Thus -v will be positive and the square root makes. sense Similarly if the object is moving upward, v> 0. the value of f(v)) is negative. The object is dropped from a height: of 500 feet with no initial velocity Using g = 32, the integral models from above becomes and where we have used the definition of f(v) for negative values of v since the object is in free fall With a little hard work or the use of a CAS, the first equation can be written as Page 4 of 5 and the second as Now if we could solve the first expression for v in terms of r and substitute the result for v in the expression for y, the result would be very unwieldy to work with. Actually it is impossible to solve the first expression to get a nice closed form for v. Thanks to technology. however, these expressions can still be used to study the object's motion Choose any value of v. say v=-1. Substituting -1 for v in the equations for t and y tells us that t = 0.032 and y = 499.984 so the point (0.032, 499.984) would lie on a graph of distance versus time Similarly, setting v = -50 provides the point (1 839, 452 462) By using a wide range of values for v. we can generate the complete graph of the distance function as a function of time Fortunately, most CAS provides the capability of plotting the points (t(v), y(v)) foe a range of values of v without having to compute each point separately Such a graph appears below. One final note. It is also possible that the integrals representing t and y cannot be integrated into a closed form No problem In this case, simply use the expressions t= 1/ - g + alpha f(V)dV y=y0+ V/ - g + alpha f(V)dV Page 5 of 5 and the numerical integrators provided by a CAS to determine values of t und y Exercises In all of the exercises you will need a CAS or related tool In the example above the graph indicates the object strikes the ground some time between 6 and 7 seconds from release Determine this time to three decimal places In the case where alpha = 1/2 ,f(v) = -v, v0 = 0, y0=100 determine closed form. expressions for v and y as function. Of t. An object is thrown upward at 40 ft/sec from a height of 100 feet The air resistance can be modeled by alpha =0.001 and f(v) = - |v| v Using the trajectory model determine the speed of the object when it strike the ground, determine the time the object strikes the ground after takeoff, construct the graph of the object's height versus time.

MAT136H1 Chapter Notes - Chapter 3b: Partial Fraction Decomposition, Antiderivative

MAT136H1 Full Course Notes

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