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PRACTICE PROBLEMS: SET 2 MATH 101: PROF. DRAGOS GHIOCA 1. Problems Problem 1. A particle travels on the x-axis with velocity given by the function v(t) = 2t − 5. (a) Find the displacement of the particle in the time interval [1, 4]. (b) Find the total distance traveled by the particle in the same time interval. Problem 2. Evaluate ∫ Problem 3. Let −3 + x2 √ dx . 5 x ∫ x4 f (x) = √ t · cos(t2 ) dt . x Compute f ′ (x). Problem 4. Compute ∫ 1 2 ex dx . x2 1 Problem 5. Let ∫ 2 sin3 (t) dt . f (x) = ex Find f ′ (x). Problem 6. Compute ∫ 6 1 2x − 1 dx . (x + 2)3 1 2 MATH 101: PROF. DRAGOS GHIOCA 2. Solutions Problem 1. (a) The displacement s(4) − s(1) is the net change in the position function s(t) for the particle in the time interval [1, 4], and it equals (by the Net Change Theorem) the integral of the velocity: ∫ ∫ 4 s(4) − s(1) = 4 (2t − 5) dt . v(t) dt = 1 1 The antiderivative for 2t − 5 is t2 − 5t and so, by the Fundamental Theorem of Calculus, we obtain that the displacement equals (t2 − 5t) |41 = (16 − 20) − (1 − 5) = −4 + 4 = 0. Hence, the particle ended exactly where it started the movement in the time interval [1, 4]. (b) The total distance traveled by the particle is the integral of the absolute value of the velocity: ∫ ∫ 4 |v(t)| dt = 1 4 |2t − 5| dt . 1 Now, we split the above integral on the intervals [1, 2.5] and [2.5, 4] since at t = 2.5, the function v(t) = 2t − 5 changes from negative to positive, and so the definition of its absolute value changes at this point. Therefore, the total distance traveled by the particle is ∫ ∫ 5 2 ∫ |2t − 5| dt + 1 = 5 2 ∫ (5 − 2t) dt + 1 4 5 2 |2t − 5| dt 4 5 2 (2t − 5) dt 5 = (5t − t2 ) |12 +(t2 − 5t) |45 2 ( ) ( ) 25 25 25 25 = − − (5 − 1) + (16 − 20) − − 2 4 4 2 25 25 = −4−4+ 4 4 9 = . 2 ∫ Problem 2. In general, the indefinite integral ∫f (x) dx represents the most general antiderivative of the function f (x); i.e., f (x) dx = F (x) + C where F ′ (x) = f (x). PRACTICE PROBLEMS: SET 2 3 We evaluate the integral by first dividing both terms in the numerator by the given denominator: ∫ −3 + x2 √ dx 5 x ∫ 1 9 = −3x− 5 + x 5 dx ∫ ∫ 9 − 15 = −3 x dx + x 5 dx 5 4 5 14 = −3 · x 5 + x 5 + C 4 14 15 4 5 14 = − x5 + x 5 + C . 4 14 Problem 3. We define two new functions: ∫ x4 √ g(x) = t cos(t2 ) dt 0 ∫ and x h(x) = √ t cos(t2 ) dt . 0 Then f (x) = g(x) − h(x), which yields that f ′ (x) = g ′ (x) − h′ (x). We compute g ′ (x) and h′ (x) using the chain rule and the Fundamental Theorem of Calculus. So, g ′ (x) = √ x4 cos((x4 )2 ) · 4x3 = x2 · cos(x8 ) · 4x3 = 4x5 cos(x8 ), and So, h′ (x) = √ x cos(x2 ). f ′ (x) = 4x5 cos(x8 ) − √ x cos(x2 ). Problem 4. First we compute the antiderivative ∫ 1 ex dx . x2 For this we use the substitution u = x1 , which yields du = − So, ∫ 1 dx . x2 1 ex dx x2 ∫ = −eu du = −eu + C 1 = −e x + C . 4 MATH 101: PROF. DRAGOS GHIOCA Therefore, using the Fundamental Theorem of Calculus, we obtain ∫ 2 1 ( 1) √ 1 ex dx = −e x |21 = −e 2 − (−e1 ) = − e + e. 2 1 x Problem 5. We have that ∫ f (x) = − ex sin3 (t) dt, 2 and so, using the Fundamental Theorem of Calculus and the chain rule, we obtain f ′ (x) = − sin3 (ex ) · (ex )′ = −ex sin3 (ex ). Problem 6. We first evaluate the indefinite integral using the substitution u = x + 2, and thus du = dx. Note that 2x − 1 = 2(x + 2) − 5 = 2u − 5. So, ∫ 2x − 1 dx (x + 2)3 ∫ = ∫ = = 2u − 5 du u3 2u−2 − 5u−3 du ∫ ∫ 2 u−2 du −5 u−3 du 5 −2 ·u +C 2 5 = −2(x + 2)−1 + (x + 2)−2 + C . 2 = −2u−1 + So, we evaluate the integral ∫ 6 2x − 1 dx 3 1 (x + 2) ( ) 5 −1 −2 = −2(x + 2) + (x + 2) |61 2 5 5 = −2 · 8−1 + 8−2 − (−2 · 3−1 + 3−2 ) 2 2 1 5 2 5 = − + + − 4 128 3 18 205 . = 1152