MATH 27 Wednesday, 5 December 208 Today: Review for Exam 3 Exam 3: Thursday, December 6; Sections 4.8-6. /6
Information on Exam 3 Six numbered problems First problem is multiple choice (five parts) See review problems online for practice! 2/6
Information on Exam 3 3/6
Integrals Ex. The graph below shows the tra c on an Internet service provider s data line from midnight to 08 : 00 AM. D is the data throughput, measured in megabits per second. What would the value of the integral s 8 D(t) dt represent? 0 The integral represents the total amount of data transmitted by the Internet service provider during this time period. 4/6
Riemann Sums Ex. Write the integral 3 dx as a limit of Riemann sums. x When writing out the Riemann sums, we need to make small rectangles. To do this, we need a length and a height of each box. The length is given by: x = b a n = 3 n = 2 n where n is the number of rectangles. Our interval endpoints are: x i = a + i x =+ 2 n i. Then, we can use right endpoints as our heights for our rectangles (you could use left endpoints or midpoints if you d like). This gives us f(x ú i ) x = 2 + 2 n i 2 n 5/6
Riemann Sums Ex. Write the integral 3 dx as a limit of Riemann sums. x f(x ú i ) x = 2 + 2 n i 2 n Now, we want to add these all up together so we get nÿ nÿ f(x ú i ) x = 2 + 2 n i 2 n i= Now we have a Riemann sum but we want to make sure we get the definition of the integral, we need to send our number of rectangles to infinity. Thus, i= 3 x dx = lim næœ i= nÿ 2 + 2 n i 2 n 6/6
Indefinite Integrals Ex. Calculate e 2x + e 22 dx. First, let us deal with A: e 2x + e 22 dx = B e 2x dx + e 2 dx A e 2x dx = e u du 2 = e u du 2 = 2 eu + C = 2 e2x + C STUFF Let u =2x du =2dx du 2 = dx 7/6
Indefinite Integrals Ex. Calculate e 2x + e 22 dx. Next, let us deal with B: e 2 dx = e 2 e 2x + e 22 dx = B e 2x dx + e 2 dx A = e 2 dx dx = e 2 x + C 2 STUFF 8/6
Indefinite Integrals Ex. Calculate e 2x + e 22 dx. Putting it all together... e 2x + e 22 dx = B e 2x dx + e 2 dx A = 2 e2x + C + e 2 x + C 2 = 2 e2x + e 2 x + C Note. Each of the two indefinite integrals generated their own constants C and C 2. Since these are both numbers getting added together here, we can just say that we get a new constant C where C = C + C 2. 9/6
Fundamental Theorem of Calculus Ex. Di erentiate the following function. g(x) = 5 x cot(t 5 ) dt. The first thing we want to realize here is that we are being asked to take the derivative of an integral. Sounds like the Fundamental Theorem of Calculus to me! The FTC says that if f is continuous on [a, b], then the function g defined by g(x) = x a f(t) dt on a Æ x Æ b is continuous on [a, b] and di erentiable on (a, b), and g Õ (x) =f(x). 0 / 6
Fundamental Theorem of Calculus Ex. Di erentiate the following function. g(x) = 5 x cot(t 5 ) dt. Before we can apply the theorem, we need to make our function look like the theorem: 5 x g(x) = cot(t 5 ) dt = cot(t 5 ) dt x 5 Now, taking the derivative by the FTC we get g Õ (x) = cot(x 5 ) / 6
Distance vs. Displacement Ex. An object moves with velocity v(t) =t 2 t, wherev is measured in meters per second. (a) Find the displacement of the object. (b) Find the total distance traveled by the object during the time interval [0, 3]. (a) The displacement doesn t care about where v(t) is positive and negative, so we get that displacement = v(t) dt = (t 2 t) dt = 3 t3 2 t2 + C where C represents the initial position. 2 / 6
Distance vs. Displacement Ex. An object moves with velocity v(t) =t 2 t, wherev is measured in meters per second. (a) Find the displacement of the object. (b) Find the total distance traveled by the object during the time interval [0, 3]. (b) The distance on the other hand cares about when you went forward and when you went backward. To do this, we need a quick sketch to see where v(t) is positive and where it is negative: v(t) is negative from 0 to and v(t) is positive from to 3. 3 / 6
Distance vs. Displacement Ex. An object moves with velocity v(t) =t 2 t, wherev is measured in meters per second. (a) Find the displacement of the object. (b) Find the total distance traveled by the object during the time interval [0, 3]. (b) Thus, distance traveled = = 0 0 = 3 ( v(t)) dt + (t t 2 ) dt + 3 v(t) dt (t 2 t) dt = 8 32 meters 4 / 6
Areas Between Curves Ex. Find the area bounded by the curves y = x and y = x 2 2. First, we want to sketch the region: From our graph we see the points of intersection are x = 2 and x = 2. Now, we can set up our integral: A = 2 2 2 x (x 2 2) dx 5 / 6
Areas Between Curves Ex. Find the area bounded by the curves y = x and y = x 2 2. A = 2 2 2 x (x 2 2) dx How do you integrate an absolute value? A = = 2 2 0 2 = = 20 3 2 x (x 2 2) dx 2 x (x 2 2) dx + 2 0 2 x (x 2 2) dx 6 / 6