Unformatted text preview: Please don’t worry if you don’t understand how the above antiderivative was obtained.
You are not supposed to at this point. Indeed, one of the main topics we will consider
in this course is how to find antiderivatives  a process which itself is also called
integration. Exercise 5 1. Use the Fundamental Theorem Calculus to evaluate the following definite
integrals. a. 1:5dx b. j:(—3)dx c. £de (where K is a constant)
d. J::(2x + 5)dx e. (3(2); + 5)dx f. Ii4(2x + 5)dx g. I: mxdx (where m is a constant) h. j:(mx + b)dx (where m and b are constants)
2. Use the antiderivative given in Example 7 above to evaluate (approximately) 1/2
I 1 —x2 dx.
0 (Note: You should only use a calculator to compute the approximate value of
arcsin(1/2).)
3. Use the antiderivative given in Example 7 above to evaluate (approximately) {1/21/1—162 aix.
1 4. Add the two answers you obtained in problems 2 and 3 above. Verify that the
answer you get is the same as the answer you get when you compute [:1’1—x2 dx. Does it make sense to you that these answers should be the same? Explain. 5. Suppose that we have a differentiable function y : F(x) and suppose that x1 and
x2 are two specific points in the domain ofF such that F(x]) : yl and F(x2) : 312.
Then “dy _
x] dxdx—‘ 6. Explain how you can tell, just by looking at the graph ofy = cos(x), that
n/Z
j cos(x)dx < l.
0 2
(Hint: Compare the given integral to the area of a certain rectangle.) 13 ...
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 Fall '08
 HOOVER
 Calculus

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