1. In Topic 4.2 (The Definite Integral II), we considered f(x)=19-x on [-3, 3] We know that the function produces a semi-circle and the definite integral gives the area between the curve and the x-axis. [, 9-x3 dx =_ 1(3) = Due to symmetry, we know that the bottom half of this circle has the same area as the top, so we can double our previous result to find the area of a whole circle with radius 3 units. 9 A =2 -1 Now think about this function on [0, 3]. We know the area given by 1 9-x' dx = 2, How can we use this information and an understanding of symmetry to find the area given by [, vo - x' dx Can we generalize this to situations beyond circles? Give an example (even if you don't know how to compute a particular area yet). 2. Let f (x) = e* on [0, 4] You will soon be able to compute AVG[0,4] = 13.4 Sketch the function and a geometric interpretation of the average-value given. Use words to describe the inherent geometry in this example.MTH252 Topic E m Denite Integral Calculus II Now that we have some of this denite integral thing (yeah, sure, let's add together an infinite number of rectangles, each with width zero, uh huh), let's see what we can do. 0 How do we use it, algebraically (how do we manipulate it)? o What shortcuts can we take? We have already seen one property of the denite integral: We can subdivide a region and the sum of the areas of the sub-regions equals the area of the whole region. For an integrable function f over an interval [a,b] with a g c E b, Efldx:I:f(x)dx+-th(x)dx To see this, consider the top half of a circle with radius 3 (centered at the origin). r I r- I I. I ,. I 1 I. I L I |. I L Note: This property may seem intuitive if a 11 c S b, it is true even if c E [(1, b]. OK, we can do that (especially when portions of the function are negative). But, really, what else can we do to the denite integral {while whistling}? What would it mean for the lower and upper limits of integration to be the same? fjxldx Because the difference in the limits of integration give the width of the region, we would have a region of zero width. That is, we would have a vertical line segment from the x-axis at x x a to the point (a, f [(1)]. As lines have length, but not area, we have J':f(x)dx:0 MTH-252 - Calculus II T.4.2: Definite Integral II p2 of 2 What happens when we swap the limits of integration? How does J" f (x) dx relate to " f (x) dx? It turns out that these are opposites (negatives), so that I" f ( x ) dx = - [" f (x) dx This seems reasonable if we consider that ['s ( x ) dx => Ar= b-a n Therefore, ['s ( x ) dx => Ar= a-b b - a n n Two more properties are derived directly from the properties of differentiation (which were derived from the properties of limits). Constant Multiple and Sum Rules [ " c. f ( x) dx = c. ['f (x)dx I ' [ s ( x ) + 8 (x) ] dx = [ f (x) dx + 1 8(x) dx We can split functions into more basic units and treat them individually when integrating (useful). Exercises Let f and g be functions such that each statement is true. J. f ( x ) dox =- 3 J. 8(x) dx = 4 ['s ( x) dx = 2 h8(x) dx = -1 These statements are true (you will be able to show them soon). fix dx = 8 ['x' dx = 4 1 x dx = 39 [ x dx = 609 Evaluate each integral. [' f ( x) dx 2. 18( x) dx 3. [[f (x) + 8(x) ]dx 4. [' (3x2 - 4x]) dx 5. [ (3x2 - 4x] ) dox 6. [[2x3 - 78(x) ] dx