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School of Mathematics and Statistics
Carleton University
Math. 69.104

ASSIGNMENT 2
SOLUTIONS

Due: November 16, 1999 (or Nov. 15, 1999 Mr. Dubé),

Solution
a) The intercepts are found as follows: y =0 when x=-1. When x=0 we have y=1.

b) tex2html_wrap_inline323 , so f is increasing when x < 1 and decreasing when x > 1.

c) tex2html_wrap_inline331 . So, the graph of f is concave up when tex2html_wrap_inline335 , that is, when tex2html_wrap_inline337 or when tex2html_wrap_inline339 . The graph is concave down in the interval tex2html_wrap_inline341

d) The points of inflection occur when there is a change in concavity. In this case they are found to be at tex2html_wrap_inline343 and at tex2html_wrap_inline345 .

e) There is only one critical point (a local maximum) at x=1 (since tex2html_wrap_inline349 ). Use your plotter to sketch the graph. You'll find:

[5] 2a)
Sketch the graph of tex2html_wrap_inline351 over the interval [-1,2].
[5] 2b)
Evaluate tex2html_wrap_inline355 and interpret your result as an area.

The integral is the area under the curve above between the lines x=-1, x=2 and above the x-axis.

Solution Use the Box method to show that

displaymath303

Now the integral breaks up into 3 pieces as follows:

displaymath365

displaymath367

[5] 3a)
Evaluate using any method: tex2html_wrap_inline369 . Solution Use the Substitution Rule: Let tex2html_wrap_inline371 so that tex2html_wrap_inline373 . The integral now looks like,

displaymath375

displaymath377

[5] 3b)
Evaluate tex2html_wrap_inline379

Solution Use the Substitution Rule: Let tex2html_wrap_inline381 so that tex2html_wrap_inline383 . The integral now looks like,

displaymath385

or

displaymath387

[5] 3c)
Evaluate tex2html_wrap_inline389

Solution Use the Substitution Rule first: Let tex2html_wrap_inline391 so that tex2html_wrap_inline393 . The integral now looks like,

displaymath395

Bring this to a more recognizable form where the Table Method may be used..., that is, let tex2html_wrap_inline397 Then, tex2html_wrap_inline399 and

displaymath401

displaymath403

displaymath405

displaymath407

displaymath409

[5] 3d)
Evaluate tex2html_wrap_inline411

Solution The denominator factors as tex2html_wrap_inline413 where the quadratic is irreducible. Thus, since the degree of the numerator is less than the degree of the denominator, we get

displaymath415

where A = 1 by the cover-up method. To get B, C we simply multiply throughout by the denominator and find

displaymath421

since A=1. Comparing coefficients we see that B=1 and C=-1. It follows that

displaymath429

Integrating we get

eqnarray226

Total: /40


next up previous
Next: About this document

Angelo Mingarelli
Thu Dec 9 15:54:50 EST 1999