alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

# Let x 0 1 a nd f c x 0 2 plot f c x over inter

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Unformatted text preview: g FC ( x) In the previous chapter we saw the relation between the area under the cur ve area curv ar of continuous f(x) over the interval [a, b] and the CHANGE in F(x) over [a, b]. continuous contin Let: F− 1 (x) = x 2 − 1 ⇒ f −1 (x) = F−’1 (x) = 2x F 0 (x) = x 2 ⇒ f 0(x) = F ’0(x) = 2 x and In all 3 cases : F+1 (x) = x 2 + 1 ⇒ f +1 (x) = F ’ (x) = 2x +1 f(x) = F ’ (x) = 2x. Also, in all 3 cases: F(x) = ANTIDERIVATIVE {f(x)} = ∫ f (x) dx = x 2 . Again, in all 3 cases the CHANGE over the inter val [a, b] is the same: F−1 (b) − F −1 (a) = F0(b) − F0(a) = F+1 (b) − F +1 (a) = b 2 − a 2 . F−1 (a) − F−1 (b) = F 0(a) − F 0(b) = F+1 (a) − F+1 (b) = a 2 − b 2 . So given only f(x) how can we compute F C (x) ? Given continuous f(x) we can find F0(x) and plot the graph of F0 (x) directly. We continuous contin saw the plot of F0(x) in the 2 examples in chapter 27. But we cannot plot the graphs of F−1(x) and F+1(x) directly. This is because we do not know the constant term. The constant − 1 in F−1(x) and + 1 in F+1(x) is known as the CONSTANT OF INTEGRATION and is denoted by C . In the case of F0(x) the constant C = 0. For each value of C we have a cur ve similar to the cur ve of F 0 (x). In fact, F C ( x) = F(x) + C is a family of cur ves. This is the Geometric v iew of the G eometric role of C , the CONSTANT OF INTEGRATION. 157 F +1(x) = x 2 + 1 =2 f(x) 3 x 4 F 0(x) = x 2 2 F −1 (x) = x 2 − 1 -- 3 -- 2 1 -- 1 0 -1 -2 -3 -4 158 1 2 3 x To know FC (x) we need to know C . To calculate C we need to know FC (x) at some point, say x0 . Then from FC (x) = F(x) + C we have : C = FC (x0) − F(x0) . In F0 (x) = x2 the constant C = 0 since F0 (x) = F(x) everywhere. If we know F(x) and C then it is easy to plot FC (x) . Suppose we do NOT know F(x) and C . We only know f(x) , x 0 a nd F C ( x 0). How can we plot FC (x) without finding the expression FC (x) ? (x) (x) If F(a) is known, then from a going LEFT to RIGHT in the positive direction to b...
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