410 - antiderivatives Determining antiderivatives or...

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Unformatted text preview: antiderivatives Determining antiderivatives or indefinite integrals is essentially "undoing differentiation." When we are to determine the antiderivative of , we are to find the functions such that . The basic integration rules on the top half of reference page 6 in the back of the text are just the undoing of the comparable differentiation rules given on the previous page. Note that an arbitrary constant is added to each integration rule because the derivative of any constant is . You need to memorize these rules, just as you did the differentiation rules. You are responsible for basic rules 1 thru 18. To be able to apply the rules, each function must be in the form shown in the list! If the expression for which you are to determine the antiderivative is not in one of these forms (EXACTLY), then you must rewrite the expression into one or more of these forms, if possible. Note that the symbol antiderivative of antiderivative of antiderivative of antiderivative: 2 8 antiderivative of antiderivative of is means antiderivative is sec is tan then we can determine the we first rewrite the function as 10 antiderivative of MTH 173 section 4-10 notes, page 1 18 here we first determine arctan arctan So 34 arctan what's ? , then use the given condition to determine a specific value of : so then so 46 which graph - antiderivative of graph is a graph of an antiderivative of is the derivative graph of MTH 173 section 4-10 notes, page 2 48 velocity graph given. draw position graph 54 draw curve direction field below 60 given determine so MTH 173 section 4-10 notes, page 3 74 A car travelling at 50 mi/hr when the brakes are applied, producing a deceleration of 22 ft/s . What distance to stop? in this problem, I use the conversion of mi hr So we are given: we know: so that So: and ft s, ft ft s, mi hr ft s to obtain consistent units: but so now to determine stopping distance, we determine stopping time from the velocity function, that is, what is ? Then we substitute that time result into the position function + s to stop s ! " ft to stop MTH 173 section 4-10 notes, page 4 ...
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