Section 5.1

Find the area under the curve f(x) = −\frac{1}{2}x + 4, above the x−axis and between x = 2 and x = 6.

Method 1: In this case our function is very simple, so we can determine the area without calculus:

Method 2: Estimate using rectangles. Inscribed rectangles with Δx = 1:

f(3)(1) + f(4)(1) + f(5)(1) + f(6)(1) =

= [−\frac{1}{2}(3) + 4](1) + [−\frac{1}{2}(4) + 4](1)

+[−\frac{1}{2}(5) + 4](1) + [−\frac{1}{2}(6) + 4](1)

= \frac{5}{2}(1) + 2(1) + \frac{3}{2}(1) + 1(1) = 7 Inscribed rectangles with Δx = \frac{1}{2}:

f(\frac{5}{2})(\frac{1}{2}) + f(3)(\frac{1}{2}) + f(\frac{7}{2})(\frac{1}{2}) + f(4)(\frac{1}{2})

+f(\frac{9}{2})(\frac{1}{2}) + f(5)(\frac{1}{2}) + f(\frac{11}{2})(\frac{1}{2}) + f(6)(\frac{1}{2})

= [−\frac{1}{2}(\frac{5}{2}) + 4](\frac{1}{2}) + [−\frac{1}{2}(3) + 4](\frac{1}{2}) + [−\frac{1}{2}(\frac{7}{2}) + 4](\frac{1}{2}) + [−\frac{1}{2}(4) + 4](\frac{1}{2}) + [−\frac{1}{2}(\frac{9}{2}) + 4](\frac{1}{2}) + [−\frac{1}{2}(5) + 4](\frac{1}{2}) + [−\frac{1}{2}(\frac{11}{2}) + 4](\frac{1}{2}) + [−\frac{1}{2}(6) + 4](\frac{1}{2}) = \frac{11}{4}(\frac{1}{2}) + \frac{5}{2}(\frac{1}{2}) + \frac{9}{4}(\frac{1}{2}) + 2(\frac{1}{2}) + \frac{7}{4}(\frac{1}{2}) + \frac{3}{2}(\frac{1}{2}) + \frac{5}{4}(\frac{1}{2}) + 1(\frac{1}{2}) = \frac{15}{2} Inscribed rectangles with Δx = \frac{6−2}{n} = \frac{4}{n}:

Circumscribed rectangles with Δx = 1: f(2)(1) + f(3)(1) + f(4)(1) + f(5)(1) = = [−\frac{1}{2}(2) + 4](1) + [−\frac{1}{2}(3) + 4](1)

+[−\frac{1}{2}(4) + 4](1) + [−\frac{1}{2}(5) + 4](1)

= 3 + \frac{5}{2}(1) + 2(1) + \frac{3}{2}(1) = 9 Defn: ∫_{a}^{b} f(x)dx = lim_{n→∞}Σ_{i=1}^{n}f(x_{i})Δx If f is continuous, can use inscribed rectangles, circumscribed rectangles, all left-hand endpoints, all right-hand endpoints, or all midpoints, etc. If Δx = \frac{b−a}{n} and if right-hand endpoints are used, then x_{i} = a + iΔx = a + \frac{(b−a)i}{n}

∫_{a}^{b} f(x)dx = lim_{n→∞}Σ_{i=1}^{n}f(a + \frac{(b−a)i}{n})(\frac{b−a}{n}) Properties of the definite integral ∫_{a}^{a}f(x)dx = 0 ∫_{a}^{b}f(x)dx = −∫_{b}^{a}f(x)dx ∫_{a}^{b}kf(x)dx = k∫_{a}^{b}f(x)dx ∫_{a}^{b}(f_{1} + f_{2})(x)dx = ∫_{a}^{b}f_{1}(x)dx + ∫_{a}^{b}f_{2}(x)dx ∫_{a}^{c}f(x)dx = ∫_{a}^{b}f(x)dx + ∫_{b}^{c}f(x)dx If f_{1}(x) ≤ f_{2}(x), then ∫_{a}^{b}f_{1}(x)dx ≤ ∫_{a}^{b}f_{2}(x)dx If m ≤ f(x) ≤ M then m(b − a) ≤ ∫_{a}^{b}f(x)dx ≤ M(b − a)

Estimate the distance traveled between t = 2 and t = 6 if the velocity is given by the function f(t) = −\frac{1}{2}t + 4. Estimate using inscribed rectangles with Δt = 1:

f(3)(1) + f(4)(1) + f(5)(1) + f(6)(1) = = [−\frac{1}{2}(3) + 4](1) + [−\frac{1}{2}(4) + 4](1)

+[−\frac{1}{2}(5) + 4](1) + [−\frac{1}{2}(6) + 4](1)

= \frac{5}{2}(1) + 2(1) + \frac{3}{2}(1) + 1(1) = 7 Find the distance traveled between t = 2 and t = 6 if the velocity is given by the function f(t) = −\frac{1}{2}t + 4.

Method 1: In this case our function is very simple, so we can determine the area without calculus:

Method 2: Use calculus by estimating with rectangles and taking limit.

Area = = lim_{n→∞}Σ_{i=1}^{n}f(a + \frac{(b−a)i}{n})(\frac{b−a}{n})

= lim_{n→∞}Σ_{i=1}^{n}f(2 + \frac{4i}{n})(\frac{4}{n})
= lim_{n→∞}Σ_{i=1}^{n}[−\frac{1}{2}(2 + \frac{4i}{n}) + 4](\frac{4}{n}) = 8

Method 3 (section 5.3): Use calculus by integrating.

∫_{2}^{6}(−\frac{1}{2}t + 4)dt = (−\frac{1}{4}t^{2} + 4t)|_{2}^{6}

= (−\frac{1}{4}(6)^{2} + 4(6)) − (−\frac{1}{4}(2)^{2} + 4(2))
= −9 + 24 − (−1 + 8) = 15 − 7 = 8