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6. Integration 6.5: Antiderivatives More Properties of Definite Integrals Ra 1. a f (x) dx = 2. Rb 3. Ra (f (x) ± g(x)) dx = a 4. Rb 5. Rb a a a cf (x) dx = f (x) dx = f (x) dx = Example 1. Find the area represented by a) Z 1 x dx. −2 b) Find the antiderivative F (x) of f (x) = x, and evalute F (1) − F (−2). 1 The Fundamental Theorem of Calculus, Part I (FTC I): If f is continuous on [a, b], then the Rx function g defined by a f (t) dt is continuous on [a, b] and differentiable on (a, b), and g 0 (x) = f (x). Example 2. Find the derivative of each function a) Z x (t2 − 1)2 dt; g(x) = 1 b) Z u g(u) = t3 1 1 dt; +1 The Fundamental Theorem of Calculus, Part II (FTC II): Suppose f is a continuous function defined on a closed interval [a, b], then Z b f (x) dx = F (b) − F (a) a where F is any antiderivative of f . Example 3. Evaluate the following integrals: a) Z 3 e3x dx −2 b) Z B√ 6 2x + 5 dx, −1 where B is a real number and B > −1 2 c) Z 4 e3x−4 dx 2 d) Find the average value of the function f (x) = 3x2 − x on the interval [1, 3] e) Z 1 2 1 dx (−x + 5)3 3 f) Z 1 0 x2 Z 5 x+3 dx + 4x + 3 g) | − x + 7| dx −5 4 6.5: Fundamental Theorem of Calculus (continue) Example 4. Find the area under the parabola y = −(x − 1)2 + 1 for x in the interval [0,2]. NOTE: BecauseR of the relation given by the FTC between the antiderivatives and integrals, the indefinite integral f (x) dx as a function is traditionally used for an antiderivative of f . The connection between definite integral (as a scalar) with the indefinite integral (as a function) is given by FTC: If f is continuous on [a, b] then b Z b Z f (x) dx f (x) dx = a . a Example 5. (Stewart) Suppose an object is moving along a straight line with position function s(t), velocity function v(t) and acceleration function a(t). Since s0 (t) = v(t), the FTC gives Z t2 v(t) dx = s(t2 ) − s(t1 ), t1 which is the displacement of the object during the time period [t1 , t2 ]. Hence, we can compute the displacement by integration if the velocity function is known. Similarly, since v 0 (t) = a(t), the FTC gives Z t2 a(t) dx = v(t2 ) − v(t1 ), t1 which is the change in velocity during the time period [t1 , t2 ]. If we want to calculate the distance traveled during the time interval, we have to consider the intervals when v(t) ≥ 0 (the object moves to the right) and also the intervals when v(t) ≤ 0 (the object moves to the left). In both cases, the distance is computed by integrating |v(t)|, the speed. Therefore, Z t2 |v(t)| dx = total distance traveled. t1 5 Example 6. An object moves along a line so that its velocity at time t is v(t) = 8t + 1 where position is given in meters and t is given in seconds. a) Find the displacement of the particle during the time period [0, 2]. b) Find the distance traveled during the this time period. Finding Approximations of Definite Integrals on the Calculator: (Bollinger) If f is a continous Rb function on [a, b], then the value a f (x) dx can be estimated by the following command M AT H → 9 : f nInt → f nInt(f (x), x, a, b). Average value of a function DEFINITION: The average value of a continuous function f on the interval [a, b] is given by 1 b−a b Z f (x) dx. a Example 7. Find the average value of the following function on the given interval. a) f (x) = 9 on [1,10]. b) f (x) = 2x2 + x − 5 on [-1,1]. 6