Chapter6 - Chapter 6 The Definite I ntegral Chapter Outline...

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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 Definition Example Antidifferentiation : The process of determining f ( x ) given f ΄( x ) If , then ( 29 x x f 2 = ( 29 . 2 x x f =
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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS , 11e – Slide #6 Finding Antiderivatives EXAMPLE EXAMPLE SOLUTION SOLUTION Find all antiderivatives of the given function. The derivative of x 9 is exactly 9 x 8 . Therefore, x 9 is an antiderivative of 9 x 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. ( 29 8 9 x x f =
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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 Antiderivatives EXAMPLE EXAMPLE SOLUTION SOLUTION Determine the following. Using the rules of indefinite integrals, we have + - dx x x x 3 1 2 2 ( 29 + - + = + - dx x dx x xdx dx x x x 3 1 2 3 1 2 2 2 + - = dx x dx x xdx 1 3 1 2 2 C x x x + + - = ln 3 1 3 2 2 3 2
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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS , 11e – Slide #11 Finding Antiderivatives EXAMPLE EXAMPLE SOLUTION 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, ( 29 x x x f + = 2 ( 29 x x x f + = 2 3 2 3 2 3 3 x x + ( 29 constant. a , 3 2 3 2 3 3 C C x x x f + + = 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. ( 29 C C C f + = + + = + + = = 1 3 1 2 3 1 3 1 2 3 1 1 3 2 3 3
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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 ( 29 . 2 3 2 3 2 3 3 + + = x x x f
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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS , 11e – Slide #13 Antiderivatives in Application EXAMPLE EXAMPLE SOLUTION SOLUTION A rock is dropped from the top of a 400-foot cliff. Its velocity at time t seconds is v ( t ) = -32 t feet per second. (a) Find s ( t ), the height of the rock above the ground at time t .
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