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Chapter 6 The Definite Integral

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Antidifferentiation Areas and Riemann Sums Definite Integrals and the Fundamental Theorem Areas in the xy-Plane Applications of the Definite Integral Chapter Outline

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§ 6.1 Antidifferentiation

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Antidifferentiation Finding Antiderivatives Theorems of Antidifferentiation The Indefinite Integral Rules of Integration Antiderivatives in Application Section Outline

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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #5 Antidifferentiation DefinitionExample Antidifferentiation: The process of determining f (x) given f ΄(x) If, then

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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #6 Finding AntiderivativesEXAMPLE SOLUTION Find all antiderivatives of the given function. The derivative of x 9 is exactly 9x 8. Therefore, x 9 is an antiderivative of 9x 8. So is x 9 + 5 and x 9 -17.2. It turns out that all antiderivatives of f (x) are of the form x 9 + C (where C is any constant) as we will see next.

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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #7 Theorems of Antidifferentiation

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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #8 The Indefinite Integral

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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #9 Rules of Integration

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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #10 Finding AntiderivativesEXAMPLE SOLUTION Determine the following. Using the rules of indefinite integrals, we have

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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #11 Finding AntiderivativesEXAMPLE SOLUTION Find the function f (x) for which and f (1) = 3. The unknown function f (x) is an antiderivative of. One antiderivative is. Therefore, by Theorem I, Now, we want the function f (x) for which f (1) = 3. So, we must use that information in our antiderivative to determine C. This is done below.

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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #12 Finding Antiderivatives So, 3 = 1 + C and therefore, C = 2. Therefore, our function is CONTINUED

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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #13 Antiderivatives in ApplicationEXAMPLE SOLUTION A rock is dropped from the top of a 400-foot cliff. Its velocity at time t seconds is v(t) = -32t feet per second. (a) Find s(t), the height of the rock above the ground at time t. (b) How long will the rock take to reach the ground? (c) What will be its velocity when it hits the ground? (a) We know that s΄(t) = v(t) = -32t and we also know that s(0) = 400. We can now use this information to find an antiderivative of v(t) for which s(0) = 400. The antiderivative of v(t) is To determine C, Therefore, C = 400. So, our antiderivative is s(t) = -16t 2 + 400.

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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #14 Antiderivatives in Application s(t) = -16t 2 + 400 CONTINUED (b) To determine how long it will take for the rock to reach the ground, we simply need to find the value of t for which the position of the rock is at height 0. In other words, we will find t for when s(t) = 0. This is the function s(t). 0 = -16t 2 + 400 Replace s(t) with 0. -400 = -16t 2 Subtract. 25 = t 2 Divide. 5 = t Take the positive square root since t ≥ 0. So, it will take 5 seconds for the rock to reach the ground.

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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #15 Antiderivatives in Application v(t) = -32t CONTINUED This is the function v(t). Replace t with 5 and solve. So, the velocity of the rock, as it hits the ground, is 160 feet per second in the downward direction (because of the minus sign). (c) To determine the velocity of the rock when it hits the ground, we will need to evaluate v(5). v(5) = -32(5) = -160

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