MATH 1210 - Section 002
Fall 2010
Nicolas Fourrier
Due on November 30, 2010
Homework 10 Name: √ 1. Estimate the area under the graph of f (x) = x from x = 0 to x = 4 using four approximating rectangles and right endpoints, and then left endpoints. Which estimate is an underestimate or an overestimate. √ √ √ √ = 1 Right endpoints: R4 = f (x1 ).1 + f (x2 ).1 + f (x3 ).1 + f (x4 ).1 = 1 + 2 + 3 + 4 = 6.146. ∆x = 4−0 4 Since f is increasing [0, 4] then √ √ on √ √ this is an overestimate. Right endpoints: L4 = f (x1 ).1 + f (x2 ).1 + f (x3 ).1 + f (x4 ).1 = 0 + 1 + 2 + 3 = 4.146. Since f is increasing on [0, 4] then this is an underestimate. 2
2. Estimate the area under the graph of f (x) = e−x from x = −2 to x = 2 using four approximating rectangles and right endpoints, and then midpoints. Which estimate is an underestimate or an overestimate. In each case sketch the curve and the rectangles. Improve your two estimates by using 8 rectangles.
4 rectangles: =1 ∆x = 2−−2 4 Right endpoints: R4 = f (−1).1 + f (0).1 + f (1).1 + f (2).1 = e−1 + 1 + e−1 + e−4 = 1.754. Midpoints: R4 = f (−1.5).1 + f (−0.5).1 + f (0.5).1 + f (1.5).1 = e−1 + 1 + e−1 + e−4 = 1.768. 8 rectangles: = 0.5 ∆x = 2−−2 8 Right endpoints: R8 = 0.5[f (−1.5) + f (−1) + f (−0.5) + f (0) + f (0.5) + f (1) + f (1.5) + f (2)] = 1.761. Due to the symmetry of the figure, we see that: Midpoints: R8 = 0.5(2)[f (0.25)+f (0.75)+f (1.25)+f (1.75)] = 1.766.
3.
Determine a region whose area is equal to the given limit. Do not evaluate the limit. Let f (x1 ) + ... + f (xn ). n X 2 2i lim (5 + )10 n→∞ n n i=1
Pn
i=1
f (xi ) =
This can be interpreted as the area of the region lying under the graph of y = (5 + x)10 on the interval [0, 2]
4. Evaluate the definite integral R2 (a) −1 (x3 − 2x)dx R5 (b) −2 6dx R1 4 (c) 0 x 5 dx R2 (d) 1 x34 dx R1 √ dx (e) 0 x−1 x R 1 x+1 (f) −1 e dx
(a)
R2
(b)
R5
(c) (d) (e) (f)
−1 −2
(x3 − 2x)dx =
3 4
6dx = 42
4 x 5 dx = 59 0 R2 3 dx = 78 1 x4 R 1 x−1 √ dx = 40 3 0 x
R1
R1 −1
ex+1 dx = e2 − 1
5. What is wrong with the equation: R1 −3 (a) −2 x−4 dx = x−3 |1−2 = − 38 R1 (b) −2 x43 dx = − x22 |2−1 = 32 In both cases, the function is not continuous on the interval [−2, 1] and [−1, 2], so the fundamental theorem of calculus can not applied. 6. The percentage of families with children that are headed by single females grew at the rate of R(t) = 0.8499t2 − 3.872t+5 (0 ≤ t ≤ 3) households/decade between 1970 (t = 0) and 2000 (t = 3). The number of such households stood at 5.6% of all families in 1970. • Find an expression giving the percentage of these households in the tth decade. • If the trend continued, estimate the percentage of these households in 2010. • What was the net increase in the percentage of these households from 1970 to 2000 ?
• 0.2833t3 − 1.936t2 + 5t + 5.6 • 12.8% • The net increase was 5.2%
7. If f (x) is the slope of a trail at a distance of x miles from the start of the trail, what does ?
R5 3
f (x)dx represents
R5 R5 The slope of the trail is the rate of change of the elevation E, so f (x) = E 0 (x). So 3 f (x)dx = 3 E 0 (x)dx = E(5) − E(3) is the change in the elevation E between 3 and 5 miles from the start of the trail. 8. The marginal cost of manufacturing x yards of a certain fabric C 0 (x) = 3 − 0.01x + 0.000006x2 (in dollar per yard). Find the increase in cost if the production level is raised from 2000 yards to 4000 yards. C(4000) − C(2000) =
R 4000 2000
C 0 (x)dx = $58, 000
√ 9. The linear density of a rod of length 4m is given by p(x) = 9 + 2 x measured inkilograms per meter, where x is measured in meters from one end of the rod. Find the total mass of the rod.
Since m0 (x) = p(x) where m is the mass of the rod, then m =
R4 0
p(x)dx =
140 3
kg.
10. Find the definite integral. Credit will be given for the details R 2 1/x (a) 1 ex2 dx R1 2 (b) 0 xe−x dx Ra √ (c) 0 x a2 − x2 dx R4 x dx (d) 0 √1+2x R e4 dx (e) e x√ln x dx
(a)
R2 1
e1/x dx x2
R1
−x2
=e−
√
e
1 (1 2
xe dx = − 1e ) 0 Ra √ (c) 0 x a2 − x2 dx = 13 a3 R4 x dx = 10 (d) 0 √1+2x 3 R e4 dx (e) e x√ln x dx = 2
(b)
11. Find the areo of region enclosed by these graphs and vertical lines (a) f (x) = x + 1, g(x) = 9 − x2 , x = −1, x = 2 (b) f (x) = 12 − x2 , g(x) = x2 − 6, x = −3, x = 3 (c) f (x) =
1 , x 2
g(x) =
1 , x2
x = 1, x = 2
(d) f (x) = x , g(x) = 4x − x2
(a) f (x) = x + 1, g(x) = 9 − x2 , x = −1, x = 2, get
R2
(9 − x2 ) − (x + 1)dx = 39 2 R 3 (b) f (x) = 12 − x2 , g(x) = x2 − 6, x = −3, x = 3, get −3 (12 − x2 ) − (x2 − 6)dx = 72 R2 (c) f (x) = x1 , g(x) = x12 , x = 1, x = 2, get 1 x1 − x12 dx = ln 2 − 21 = 0.19 R2 (d) f (x) = x2 , g(x) = 4x − x2 , get 0 4x − x2 − x2 dx = 38 −1
