6-2 definite integrals

*Yesterday, we explored Riemann Sums using both a set width for the rectangles as well as varying widths. We found that if we averaged our individual answers, we were able to come close to “the real thing.” *Imagine if we were to make the widths super small – creating LOTS of rectangles. We would be practically finding the real area instead of estimating the area. Definite Integral is a limit of Riemann Sums.

Thm: All Continuous Functions are Integrable Definite Integral: height base partitions Thm: All Continuous Functions are Integrable If f is continuous on [a, b], then the definite integral exists on [a, b]. (watch how the notation morphs…) *Because the limits are all the same, we don’t need the partitions… * f is continuous on [a, b] * n subintervals upper limit variable of integration function/integrand lower limit

Ex 1) The interval [–1, 3] is partitioned into n subintervals of equal length . Let mk denote the midpoint of the kth subinterval. Express the limit Since the points were chosen from the subintervals of the partition, it’s a limit of Riemann sums. (didn’t “have” to be midpoints)

Def: Definite Integral and Area Under a Curve If y = f (x) is nonnegative and integrable on [a, b], then the area under the curve as defined using a definite integral is Integral Area Ex 2) Evaluate the integral Means area

*Functions can be above or below the x-axis…  below the x-axis gives us a negative value for area because the “height” is a negative value If we want TOTAL AREA (above AND below) = = area above axis – area below axis Thm: Integral of a Constant If f (x) = c is a constant function on [a, b], then Why does this make sense? It’s a rectangle!

Calculator Time!! MATH  9: fnInt (Enter whichever applies to your calculator) fnInt (f (x), x, a, b) OR Ex) Evaluate the following using your calculator. a) b)  (Let’s go back now and finish the last page of yesterday’s activity!)

Give a geometric explanation why some of the answers are negative and some are zero. Direction L  R + Graph above + 1 + 6) answer 7) –2 L  R + below – – 00

– 8) Area above = area below  cancel each other 9) –2 R  L – above + Area above = area below  cancel each other 9) –2 R  L – above + – 00

10) 2 R  L – below – + Same 11) 2

homework Pg. 286 #1–7, 10, 13, 14, 15, 22, 25, 31–34