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MATH6037: Partial Fractions Exercise Solutions

February 19, 2011 in MATH6037 | Leave a comment

Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from p.48 of these notes.

Question 1

(i)

Multiplying the first and last coefficients: 1\times 5=5. Now the factors of 5 are 1\times 5. So I want to rewrite the middle term -4x in terms of these:

x^2-4x-5=x^2+x-5x-5=x(x+1)-5(x+1)=(x+1)(x-5).

(ii)

Taking out the common factor x: x(x-2).

(iii)

Multiplying the first and last coefficients: 15\times6=90. The factors of 90 are \{90,1\},\{45,2\},\{30,3\},\{18,5\},\{15,6\},\{10,9\}. I want to rewrite the middle term x in terms of one of these factors:

15x^2+10x-9x-6=5x(3x+2)-3(3x+2)=(3x+2)(5x-3).

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MATH 6037: Integration by Parts Exercise Solutions

February 13, 2011 in MATH6037 | 2 comments

Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from p.35 of https://jpmccarthymaths.wordpress.com/2011/02/01/math6037-general-information/

These questions all require integration by parts although they don’t explicitly say so.

Question 1

We must choose a u and dv — where we choose u according to the LIATE rule. Here there are no logs, no inverses but there is an algebraic (sum of powers of x). Hence let u=x, dv=\cos x\,dx (find v by integrating: \int \,dv=v):

\frac{du}{dx}=1\Rightarrow du=dx; and v=\sin x.

Hence by the integration by parts formula:

I=x\sin x-\int\sin x\,dx=x\sin x-(-\cos x)+c

=x\sin x+\cos x+c

We check our solution by differentiating I (using the product rule):

\frac{d}{dx}I=x\cos x+\sin x-\sin x=x\cos x,

as required.

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MATH6037: Review of Integration Exercise Solutions

February 10, 2011 in MATH6037 | Leave a comment

Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from https://jpmccarthymaths.wordpress.com/2011/02/01/math6037-general-information/

Question 1

(a)

Now:

\frac{d}{dx}x^2=2x and \frac{d}{dx}5x=5,

hence:

\int_0^1 (2x+5)\,dx=\left[x^2+5x\right]_0^1 =(1^2+5(1))-(0^2+5(0))=6

(b)

There is no obvious anti-derivative and there is no obvious manipulation hence we are looking for the function-dervative pattern to make a substitution (or else use the LIATE rule). Notice that the top is the derivative of the bottom. Hence we will let u be the function; i.e. the bottom:

Let u=x^2+5x+1:

\frac{du}{dx}=2x+5

\Rightarrow dx=\frac{du}{2x+5}

Now put everything back into the integrand, suppressing the limits:

I=\int \frac{2x+5}{u}\,\frac{du}{2x+5}

=\int \frac{du}{u}=\ln u =[\ln (x^2+5x+1)]_0^1

=\ln(1^2+5(1)+1)-\ln(0^2+5(0)+1)

Now, what do we have to raise e to, to get 1? i.e. e^?=1; well e^0=1. Hence \ln 1=0:

I=\ln 7.

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MATH6037: General Information

February 1, 2011 in MATH6037 | 3 comments

Module Description:

MATH6037

More detailed General Information on this module (* not all one hundred percent accurate at time of publication) may be found after the table of contents in this set of incomplete notes:

MATH6037 Lecture Notes

(last updated 04 May. )

The Chain Rule

October 26, 2010 in General, Grinds, Leaving Cert, MA 1008, MATH6037, MS 2001 | 1 comment

MATH6037 please skip to the end of this entry.

The sum, product and quotient rules show us how to differentiate a great many different functions from the reals to the reals. However some functions, such as f(x)=\sin 2x are a composition of functions, and these rules don’t tell us what the derivative of \sin 2x is. There is, however, a theorem called the chain rule that tells us how to differentiate these functions. Here we present the proof. In class we won’t prove this assertion but we will make one attempt to explain why it takes the form it does. In general only practise can make you proficient in the use of the chain rule. See http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/exercise3.pdf or any other textbook (such as a LC text book) with exercises.

Proposition 4.1.8 (Chain Rule)

Let f, g:\mathbb{R}\rightarrow\mathbb{R} be functions, and let F denote the composition F=g\circ f (that is F(x)=g(f(x)) for each x\in\mathbb{R}). If a\in\mathbb{R} such that f is differentiable at a and g is differentiable at f(a), then F is differentiable at a with

F'(a)=g'(f(a))f'(a)

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