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MATH 151 Engineering Math I, Spring 2014 JD Kim Week13 Section 5.7, 6.1 Section 5.7 Antiderivatives Definition Antiderivative We call F (x) an antiderivative of f (x) if F ′ (x) = f (x). d 2 (x ) = 2x. ex) x2 is an antiderivative of 2x, because dx Definition If F is an antiderivative of f , then the most general antiderivative of f is F (x)+C, where C is any arbitrary constant. Examples) f (x) = k ⇒ f (x) = xn , if n 6= −1 f (x) = x−1 f (x) = ex f (x) = ax f (x) = cos x f (x) = sin x F (x) = kx + C ⇒ ⇒ xn+1 +C n+1 F (x) = ln |x| + C F (x) = ex + C ⇒ ax +C ln a F (x) = sin x + C ⇒ F (x) = ⇒ ⇒ F (x) = F (x) = − cos x + C 1 f (x) = sec x tan x f (x) = sec2 x f (x) = csc2 x f (x) = csc x cot x ⇒ ⇒ F (x) = sec x + C F (x) = tan x + C ⇒ F (x) = − cot x + C ⇒ F (x) = − csc x + C 1 ⇒ F (x) = arctan x + C x2 + 1 1 f (x) = √ ⇒ F (x) = arcsin x + C 1 − x2 f (x) = Ex1) Find the most general antiderivative 1-1) f (x) = x3 − 4x2 + 17 1-2) f (x) = 1-3) f (x) = √ 3 x2 − √ x3 x + x2 − 1 x3 1-4) f (x) = ex + √ 4 + 5(1 − x2 )−1/2 2 1−x 2 Ex2) Find f (x) given that f ′′ (x) = 3ex + 4 sin x, f (0) = 1 and f ′ (0) = 2 3 Ex3) Given the graph of f passes through the point (1,6) and the slope of its tangent line at (x, f (x)) is 2x + 1, find f (2). Ex4) A particle is moving according to acceleration a(t) = 3t + 8. Find the position, s(t), of the object at time t if we know s(0) = 1 and v(0) = −2. 4 Ex5) A stone is thrown downward from a 450m tall building at a speed of 5 meters per second. Derive a formula for the distance of the stone above ground level. Ex6) Suppose the acceleration of an object at time t is given by a(t) = −3j. Find the position vector function, r(t), if it is known that v(2) = i − j and r(0) = 0. 5 Ex7) Suppose that g is differentiable on (0, ∞), that x2 g ′ (x) = 2x2 + 3x for every x > 0, and that g(e) = 1. Determine g. 6 Ex8) A car braked with a constant deceleration of 40 feet per second squared, producing skid marks measuring 160 feet before coming to a stop. How fast was the car traveling when the brakes were first applied? 7 Section 6.1 Sigma Notation Definition If m and n are a non negative integers, with m < n, then n X i=m ai = am + am+1 + am+2 + · · · + an−1 + an . We call i the index of the sum, m the lower limit of the sum, and n the upper limit of the sum. 10 P ex) n = 1 + 2 + 3 + · · · + 9 + 10 = 55 n=1 6 P Ex9) 1 i=1 i + 1 Ex10) 8 P xk k=5 Ex11) Write 1 − 1 1 1 1 1 + − + − in sigma notation. 4 9 16 25 36 8 Theorem n n P P ai cai = c 1. i=m i=m 2. n P (ai ± bi ) = i=m 3. n P n P i=m ai ± n P bi i=m 1=n i=1 4. n P (c) = cn i=1 5. n P i=1 n P i= n(n + 1) 2 n(n + 1)(2n + 1) 6 i=1 2 n P n(n + 1) 3 7. i = 2 i=1 6. i2 = 9 Ex12) Evaluate 12-1) 50 P 3 i=1 12-2) 25 P 7 i=3 12-3) 100 P 6i i=1 12-4) 99 P i=3 1 1 − i i+1 n 1 P 12-5) lim n→∞ i=1 n " # 3 i +1 n 10 Ex13) If 4 P i=1 f (i) = 3 and 4 P g(i) = 13, find i=1 4 P (2f (i) + g(i)). i=1 11