121616949-math.162 - the beginning of the interval namely...

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148 Chapter 7 Integration Proof sketch for Theorem 7.4 . We want to compute G ( x ), so we start with the definition of the derivative in terms of a limit: G ( x ) = lim Δ x 0 G ( x + Δ x ) - G ( x ) Δ x = lim Δ x 0 1 Δ x ˆ Z x x a f ( t ) dt - Z x a f ( t ) dt ! = lim Δ x 0 1 Δ x ˆ Z x a f ( t ) dt + Z x x x f ( t ) dt - Z x a f ( t ) dt ! = lim Δ x 0 1 Δ x Z x x x f ( t ) dt. Now we need to know something about Z x x x f ( t ) dt when Δ x is small; in fact, it is very close to Δ xf ( x ), but we will not prove this. Once again, it is easy to believe this is true by thinking of our two applications: The integral Z x x x f ( t ) dt can be interpreted as the distance traveled by an object over a very short interval of time. Over a sufficiently short period of time, the speed of the object will not change very much, so the distance traveled will be approximately the length of time multiplied by the speed at
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Unformatted text preview: the beginning of the interval, namely, Δ xf ( x ). Alternately, the integral may be interpreted as the area under the curve between x and x + Δ x . When Δ x is very small, this will be very close to the area of the rectangle with base Δ x and height f ( x ); again this is Δ xf ( x ). If we accept this, we may proceed: lim Δ x → 1 Δ x Z x +Δ x x f ( t ) dt = lim Δ x → Δ xf ( x ) Δ x = f ( x ) , which is what we wanted to show. It is still true that we are depending on an interpretation of the integral to justify the argument, but we have isolated this part of the argument into two facts that are not too...
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