Chapter 6 The Definite Integral

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

§ 6.1 Antidifferentiation

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

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 Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #5

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #6 Finding Antiderivatives EXAMPLE Find all antiderivatives of the given function. SOLUTION The derivative of x9 is exactly 9x8. Therefore, x9 is an antiderivative of 9x8. So is x9 + 5 and x9 -17.2. It turns out that all antiderivatives of f (x) are of the form x9 + C (where C is any constant) as we will see next. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #6

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

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

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

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

Finding Antiderivatives EXAMPLE Find the function f (x) for which and f (1) = 3. SOLUTION 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. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #11

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

Antiderivatives in Application EXAMPLE 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? SOLUTION (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) = -16t2 + 400. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #13

Antiderivatives in Application 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. s(t) = -16t2 + 400 This is the function s(t). 0 = -16t2 + 400 Replace s(t) with 0. -400 = -16t2 Subtract. 25 = t2 Divide. 5 = t Take the positive square root since t ≥ 0. So, it will take 5 seconds for the rock to reach the ground. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #14

Antiderivatives in Application CONTINUED (c) To determine the velocity of the rock when it hits the ground, we will need to evaluate v(5). v(t) = -32t This is the function v(t). v(5) = -32(5) = -160 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). Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #15