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7.1 Workshop Exercises 1

For more worked examples, see Gilbert and Jordan Guide to Mathematical Methods, pages 153–164.

True or False?

i) If a real polynomial of degree nn has ss real roots, then (n-s)(n-s) is even.

ii) The degree of a real polynomial is equal to the number of distinct roots.

iii) The quadratic polynomial -2⁢x2+6⁢x-7-2x^{2}+6x-7 is irreducible over the real numbers.

iv) If x=sin⁡tx=\sin t and 0≤t≤π0\leq t\leq\pi then 1-x2=cos⁡t\sqrt{1-x^{2}}=\cos t.

See also Exercises W5.9-5.14 of MATH101. Try not to spend the whole workshop on partial fractions, for which there are lots of examples in Exercises W1.1-1.6. In particular, you should look at W1.7-1.10 on integration by substitution.

W1.1.

Express the following rational functions as the sum of a polynomial and a term f⁢(x)g⁢(x)\frac{f(x)}{g(x)} where deg⁡f<deg⁡g\deg f<\deg g:

i)⁢x-2x+4; ii)⁢x2+xx-3; iii)⁢x2-1x2+x+3; iv)⁢x3-2⁢x-1x2+2⁢x+2.\mbox{i)}\;\frac{x-2}{x+4};\;\;\mbox{ii)}\;\frac{x^{2}+x}{x-3};\;\;\mbox{iii)}% \;\frac{x^{2}-1}{x^{2}+x+3};\;\;\mbox{iv)}\;\frac{x^{3}-2x-1}{x^{2}+2x+2}.

W1.2.

Express the following rational functions as partial fractions, plus a polynomial if necessary:

i)⁢5⁢x-3x2-x; ii)⁢6⁢x-9x2-x-2; iii)⁢x2+5⁢x+5x3+4⁢x2+5⁢x+2; iv)⁢2⁢x-20x3-2⁢x2+4⁢x-8; v)⁢x3+3⁢x2-x-14x3+3⁢x2+2⁢x-6.\mbox{i)}\;\frac{5x-3}{x^{2}-x};\;\;\mbox{ii)}\;\frac{6x-9}{x^{2}-x-2};\;\;% \mbox{iii)}\;\frac{x^{2}+5x+5}{x^{3}+4x^{2}+5x+2};\;\;\mbox{iv)}\;\frac{2x-20}% {x^{3}-2x^{2}+4x-8};\;\;\mbox{v)}\;\frac{x^{3}+3x^{2}-x-14}{x^{3}+3x^{2}+2x-6}.

W1.3.

Integrate the following functions:

i)⁢x+3(x-1)2; ii)⁢1x2+9⁢iii)⁢4⁢x+9x2+1; iv)⁢x+2x2+3.\mbox{i)}\;\frac{x+3}{(x-1)^{2}};\;\;\mbox{ii)}\;\frac{1}{x^{2}+9}\;\;\mbox{% iii)}\;\frac{4x+9}{x^{2}+1};\;\;\mbox{iv)}\;\frac{x+2}{x^{2}+3}.

W1.4.

a) Find the indefinite integral of the rational functions in W1.2(i)-(iv).

b) Find the indefinite integral of the rational function in W1.2(v), using the substitution u=x2+4⁢x+6u=x^{2}+4x+6 to integrate the part of the form A⁢x+Bx2+4⁢x+6\frac{Ax+B}{x^{2}+4x+6}.

W1.5. Using partial fractions, find the following indefinite integrals:

i)⁢∫2⁢x2+7x3-x2-8⁢x+12⁢d⁢x, ii)⁢∫x3-2x3-x2+4⁢x-4⁢d⁢x.\mbox{i)}\;\int{{2x^{2}+7}\over{x^{3}-x^{2}-8x+12}}\,dx,\;\;\mbox{ii)}\;\int{{% x^{3}-2}\over{x^{3}-x^{2}+4x-4}}dx.

W1.6. i) Given that 1+3⁢i1+3{\rm i} is a root of the following equation, find all of the roots:

z4-6⁢z3+26⁢z2-56⁢z+80=0,z^{4}-6z^{3}+26z^{2}-56z+80=0,

ii) Factorize f⁢(x)=x4-6⁢x3+26⁢x2-56⁢x+80f(x)=x^{4}-6x^{3}+26x^{2}-56x+80 as a product of irreducible real quadratic polynomials.

iii) Use partial fractions to express 26f⁢(x)\frac{26}{f(x)} as a sum of expressions of the form α⁢x+βQ⁢(x)\frac{\alpha x+\beta}{Q(x)} where Q⁢(x)Q(x) is an irreducible quadratic polynomial.

iv) To integrate a rational function of the form α⁢x+βQ⁢(x)\frac{\alpha x+\beta}{Q(x)} where Q⁢(x)=x2+2⁢b⁢x+cQ(x)=x^{2}+2bx+c is an irreducible quadratic, first substitute s=x+bs=x+b so that Q⁢(x)=s2+(c-b2)Q(x)=s^{2}+(c-b^{2}) where c-b2>0c-b^{2}>0. The resulting rational function in ss can be integrated using the method which is explained in 1.16-17 in the notes. (See also the examples in 1.21-24.)

Using the result from (iii), determine the indefinite integral ∫26f⁢(x)⁢d⁢x\int\!\frac{26}{f(x)}\,dx.

W1.7. By using the substitution t=tan⁡x2t=\tan\frac{x}{2}, find the definite integral

∫0π2d⁢x5+4⁢cos⁡x.\int_{0}^{\frac{\pi}{2}}{{dx}\over{5+4\cos x}}.

W1.8. Choose substitutions to determine the following integrals:

i)⁢∫(2⁢x+3)32⁢d⁢x; ii)⁢∫d⁢xx2+4; iii)⁢∫2⁢x⁢d⁢xx2-4.\mbox{i)}\;\int(2x+3)^{\frac{3}{2}}\,dx;\;\;\mbox{ii)}\;\int\frac{dx}{\sqrt{x^% {2}+4}};\;\;\mbox{iii)}\;\int\frac{2x\,dx}{\sqrt{x^{2}-4}}.

W1.9. Use suitable substitutions to find the integrals:

i)⁢∫03d⁢x1+x2; ii)⁢∫1/31d⁢x(1+x2)2; iii)⁢∫13d⁢x(1+x)⁢x1/2; iv)⁢∫01d⁢x4-x2; v)⁢∫abx2-1⁢d⁢x\mbox{i)}\;\int_{0}^{\sqrt{3}}{{dx}\over{1+x^{2}}};\;\;\mbox{ii)}\;\int_{{1}/{% \sqrt{3}}}^{1}{{dx}\over{(1+x^{2})^{2}}};\;\;\mbox{iii)}\;\int_{1}^{3}{{dx}% \over{(1+x)x^{1/2}}};\;\;\mbox{iv)}\int_{0}^{1}{{dx}\over{\sqrt{4-x^{2}}}};\;% \;\mbox{v)}\;\int_{a}^{b}\sqrt{x^{2}-1}\,dx

where 1<a<b1<a<b in part (v).

W1.10. Use the substitution x-2=2⁢sinh⁡tx-2=2\sinh t to show that

∫02d⁢x(x2-4⁢x+8)1/2=log⁡(1+2);\int_{0}^{2}{{dx}\over{(x^{2}-4x+8)^{1/2}}}=\log(1+\sqrt{2});

recall that sinh-1⁡y=log⁡(y+y2+1).\sinh^{-1}y=\log(y+\sqrt{y^{2}+1}).