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5.2 Definite Integrals

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jetta
21 diapositives · 398 Views · Sep 11, 2014

This text explores the concept of definite integrals in calculus, detailing how the area under a curve is approximated using Riemann sums and subinterval partitions. It explains the definitions of integrals, the importance of dummy variables, and discusses the role of continuous functions in integration. The text illustrates the connection between velocity, time, and distance, establishing that distance can be formulated as the area under a curve. Additionally, it demonstrates how to evaluate integrals graphically and algebraically, emphasizing areas under curves involving trapezoids, and considers both positive and negative areas.

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5.2 Definite Integrals
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  1. 5.2 Definite Integrals

  2. When we find the area under a curve by adding rectangles, the answer is called a Riemann sum. The width of a rectangle is called a subinterval. The entire interval is called the partition. subinterval partition Subintervals do not all have to be the same size.

  3. If the partition is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by . As gets smaller, the approximation for the area gets better. subinterval partition if P is a partition of the interval

  4. is called the definite integral of over . If we use subintervals of equal length, then the length of a subinterval is: The definite integral is then given by:

  5. Leibniz introduced a simpler notation for the definite integral: Note that the very small change in x becomes dx.

  6. It is called a dummy variable because the answer does not depend on the variable chosen. upper limit of integration Integration Symbol integrand variable of integration (dummy variable) lower limit of integration

  7. Note: all continuous functions are integrable. That is, if a Function is continuous on [a,b] then its definite integral over [a,b] exists. We have the notation for integration, but we still need to learn how to evaluate the integral.

  8. velocity time In section 5.1, we considered an object moving at a constant rate of 3 ft/sec. Since rate . time = distance: If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line. After 4 seconds, the object has gone 12 feet.

  9. If the velocity varies: Distance: (C=0 since s=0 at t=0) After 4 seconds: The distance is still equal to the area under the curve! Notice that the area is a trapezoid That can be solved graphically.

  10. What if: We could split the area under the curve into a lot of thin trapezoids, and each trapezoid would behave like the large one in the previous example. We would use the sum expressions to solve. It seems reasonable that the distance travelled will again equal the area under the curve.

  11. The area under the curve We can use anti-derivatives to find the area under a curve!

  12. Note – for some problems, using algebraic integration is either very difficult or even impossible! – drawing a diagram and solving pictorially is still an excellent method! Use the graph of the integrand and areas to evaluate the integral: Since this is really asking for the area of a quarter circle of radius 5, we can use the area formula! Area

  13. Let area under the curve from a to x. (“a” is a constant) Let’s look at it another way: Then:

  14. min f max f h The area of a rectangle drawn under the curve would be less than the actual area under the curve. The area of a rectangle drawn above the curve would be more than the actual area under the curve.

  15. As h gets smaller, min f and max f get closer together. This is the definition of derivative! initial value Take the anti-derivative of both sides to find an explicit formula for area.

  16. As h gets smaller, min f and max f get closer together. Area under curve from a to x = antiderivative at x minus antiderivative at a.

  17. Area = F(b) – F(a)

  18. Example: Find the area under the curve from x=1 to x=2. = = = Area under the curve from x=1 to x=2. Area from x=0 to x=2 Area from x=0 to x=1

  19. ENTER Example: Find the area under the curve from x=1 to x=2. To do the same problem on the TI-83: fnInt(x^2,x,1,2) Math 9

  20. Example: Find the area between the x-axis and the curve from to . pos. neg. On the TI-83: p fnInt(abs(cos(x)),x,0,3 /2) If you use the absolute value function, you don’t need to find the roots. p

  21. In general, if a function f takes on both positive and negative values, then we can interpret the integral as the difference of areas or the net area of the region. Where A1 is the region above the x-axis and below the graph of f Where A2 is the region below the x-axis and above the graph of f A1 A2