By Garrett P.

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**Additional resources for Calculus Refresher, version 2008.4 (2008)(en)(78s)**

**Sample text**

Of course, this isn’t exactly right, because integrals are not exactly areas, but thinking of area is a reasonable heuristic. Of course, an approximation is not very valuable unless there is an estimate for the error, in other words, an idea of the tolerance. b Each of the approaches starts the same way: To approximate a f (x) dx, break the interval [a, b] into smaller subintervals [x0 , x1 ], [x1 , x2 ], . . , [xn−2 , xn−1 ], [xn−1 , xn ] each of the same length ∆x = b−a n and where x0 = a and xn = b.

And the use of language should probably be taken to mean that at time t = 0 there are 1000 llamas, and at time t = 4 there are 2000. Then, either repeating the method above or plugging into the formula derived by the method, we find c = value of f at t = 0 = 1000 k= ln 1000 − ln 2000 ln f (t1 ) − ln f (t2 ) = t1 − t2 0−4 = ln ln 21 1000 −4 = = (ln 2)/4 2000 −4 Therefore, f (t) = 1000 e ln 2 4 t = 1000 · 2t/4 This is the desired formula for the number of llamas at arbitrary time t. Example: A colony of bacteria is growing exponentially.

Integration by parts Strangely, the subtlest standard method is just the product rule run backwards. This is called integration by parts. (This might seem strange because often people find the chain rule for differentiation harder to get a grip on than the product rule). One way of writing the integration by parts rule is f (x) · g (x) dx = f (x)g(x) − f (x) · g(x) dx Sometimes this is written another way: if we use the notation that for a function u of x, du = du dx dx then for two functions u, v of x the rule is u dv = uv − v du Yes, it is hard to see how this might be helpful, but it is.