Wed Dec 3 2014 11:59 PM AKST - University of Alaska...

Question 12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152

E5 (6607623)

Due: Wed Dec 3 2014 11:59 PM AKST

1. Question Details SCalc7 4.1.003.MI. [1874320]

(a) Estimate the area under the graph of f(x) = 5 cos(x) from x = 0 to x = π/2 using four approximating rectangles andright endpoints. (Round your answers to four decimal places.)R4 =

Sketch the graph and the rectangles.

Is your estimate an underestimate or an overestimate?

(b) Repeat part (a) using left endpoints.L4 =

Sketch the graph and the rectangles.

underestimate

overestimate

Is your estimate an underestimate or an overestimate?

Solution or Explanation

Click to View Solution

underestimate

overestimate

2. Question Details SCalc7 4.1.005. [2524211]

(a) Estimate the area under the graph of using three rectangles and right endpoints.R3 =

Then improve your estimate by using six rectangles.R6 =

Sketch the curve and the approximating rectangles for R3.

f(x) = 8 + 2x2 from x = −1 to x = 2

Sketch the curve and the approximating rectangles for R6.

(b) Repeat part (a) using left endpoints.L3=L6=

Sketch the curve and the approximating rectangles for L3.

Sketch the curve and the approximating rectangles for L6.

(c) Repeat part (a) using midpoints.M3=M6=

Sketch the curve and the approximating rectangles for M3.

Sketch the curve and the approximating rectangles for M6.

(d) From your sketches in parts (a)(c), which appears to be the best estimate?

Speedometer readings for a motorcycle at 12second intervals are given in the table.

t (s) 0 12 24 36 48 60

v (ft/s) 30 27 25 22 24 28

(a) Estimate the distance traveled by the motorcycle during this time period using the velocities at the beginning ofthe time intervals.

(b) Give another estimate using the velocities at the end of the time periods. ft

(c) Are your estimates in parts (a) and (b) upper and lower estimates? Explain.

(a) and (b) are neither lower nor upper estimates since v is neither an increasing nor decreasingfunction of t.

(b) is a lower estimate and (a) is an upper estimate since v is a decreasing function of t.

(a) is a lower estimate and (b) is an upper estimate since v is an increasing function of t.

Use the Definition to find an expression for the area under the graph of f as a limit. Do not evaluate the limit.

f(x) = x2 + , 3 ≤ x ≤ 51 + 2x

lim n → ∞

f(x) = x2 + , 3 ≤ x ≤ 5.1 + 2x Δx = (5 − 3)/n = 2/n and xi = 3 + iΔx = 3 + 2i/n.

A = Rn = = lim n → ∞

lim n → ∞

f(xi)Δxi = 1

lim n → ∞

n(3 + 2i/n)2 + · .

i = 11 + 2(3 + 2i/n) 2

Determine a region whose area is equal to the given limit. Do not evaluate the limit.

lim n → ∞

y = (9 + x)16 on [9, 11]

y = (9 + x)11 on [0, 2]

y = (9 + x)11 on [9, 11]

y = x12 on [0, 2]

y = (9 + x)12 on [0, 2]

Determine a region whose area is equal to the given limit. Do not evaluate the limit.

lim n → ∞

x tan(x) on 0, π8

x tan(x) on − , π8

tan(x) on [0, 8π]

tan(x) on 0, π8

tan(x) on − , π8

(a) Let An be the area of a polygon with n equal sides inscribed in a circle with radius r.

By dividing the polygon into n congruent triangles with central angle show that the following is true.

(b) Show that [Hint: Use the equation.]

(a) The diagram shows one of the n congruent triangles, Δ with central angle O is the center of the circle andAB is one of the sides of the polygon. Radius OC is drawn so as to bisect ∠AOB. It follows that OC intersects AB at rightangles and bisects AB. Thus, ΔAOB is divided into two right triangles with legs of length

ΔAOB has area

(b) To use the equation, we need to have the same expression in the denominator as we have in the

argument of the sine function—in this case, 2π/n.

2π/n,

An = nr2 sin12

An = πr2.lim n → ∞

AOB, 2π/n.

(AB) = r sin(π/n) and r cos(π/n).12

2 · [r sin(π/n)][r cos(π/n)] = r2 sin(π/n) cos π/n) = r2 sin(2π/n),12

An = n · area(ΔAOB) = nr2 sin(2π/n).12

= 1,lim θ → 0

sin θθ

An = nr2 πr2 = πr2 = (1) πr2 = πr2.lim n → ∞

lim n → ∞

sin(2π/n)2π/n

lim θ → 0

sin θθ

8. Question Details SCalc7 4.1.501.XP. [1874565]

The area A of the region S that lies under the graph of the continuous function is the limit of the sum of the areas ofapproximating rectangles.

Use this definition to find an expression for the area under the graph of f as a limit. Do not evaluate the limit.

A = Rn = [f(x1)Δx + f(x2)Δx + . . . + f(xn)Δx]lim n → ∞

lim n → ∞

f(x) = , 1 ≤ x ≤ 19x5

A = lim n → ∞

9. Question Details SCalc7 4.1.JIT.003. [1805435]

Find the sum.

2k − 1k = 1

The table gives the values of a function obtained from an experiment. Use them to estimate using three equal

subintervals with right endpoints, left endpoints, and midpoints.

x 3 4 5 6 7 8 9

f(x) −3.5 −2.3 −0.6 0.3 0.7 1.5 1.8

(a) Estimate using three equal subintervals with right endpoints.

If the function is known to be an increasing function, can you say whether your estimate is less than or greaterthan the exact value of the integral?

(b) Estimate using three equal subintervals with left endpoints.

(c) Estimate using three equal subintervals with midpoints.

f(x) dx9

less than

greater than

one cannot say

f(x) dx9

less than

greater than

one cannot say

f(x) dx9

less than

greater than

one cannot say

Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal places.

so the endpoints are 0, 24, 48, 72, and 96, and the midpoints are 12, 36, 60, and 84. The Midpoint

Rule gives

sin dx, n = 496

Δx = (96 − 0)/4 = 24,

sin dx ≈ 96

x4f(xi)Δx = 24(sin + sin + sin + sin ) ≈ 24(0.6545) = 15.7087.

i = 112 36 60 84

The graph of g consists of two straight lines and a semicircle. Use it to evaluate each integral.

g(x) dx10

g(x) dx30

g(x) dx35

Evaluate the integral by interpreting it in terms of areas.

2 + dx0

81 − x2−9

Evaluate the integral by interpreting it in terms of areas.

|x − 3| dx6

Evaluate

sin5 x cos5 x dx.π

Given that what is

= 15 − 24,9x dx1

x2 + 40

5 ?9u du0

u2 + 41

If where f is the function whose graph is given, which of the following values is largest?

so F(0) is negative, and similarly, so is F(4). F(12) and F(16) are negative since they

represent negatives of areas below the xaxis. Since is the only nonnegative value, F(8) is the

largest.

F(x) = ,f(t) dtx

F(0) = = − f(t) dt0

8f(t) dt,8

F(8) = f(t) dt = 08

Each of the regions A, B, and C bounded by the graph of f and the xaxis has area 3. Find the value of

= −8 + 2 = −6

[f(x) + 2x + 3] dx.2

I = = + 2 + = I1 + 2I2 + I3[f(x) + 2x + 3] dx2

−4f(x) dx2

−4x dx2

−43 dx2

−4I1 = −3 [area below xaxis] + 3 − 3 = −3

I2 = − (4)(4) [area of triangle, see figure] + (2)(2)12

I3 = 3 [2 − (−4)] = 3(6) = 18I = −3 + 2(−6) + 18 = 3.

Suppose f has absolute minimum value m and absolute maximum value M. Between what two values must lie?

(smaller value)

(larger value)

Which property of integrals allows you to make your conclusion?

f(x) dx4

If f(x) ≥ 0 for a ≤ x ≤ b, then ≥ 0.f(x) dxb

If m ≤ f(x) ≤ M for a ≤ x ≤ b, then m(b − a) ≤ ≤ M(b − a).f(x) dxb

= 0f(x) dxa

+ = f(x) dxc

af(x) dxb

cf(x) dxb

= − f(x) dxb

af(x) dxa

If m ≤ f(x) ≤ M for a ≤ x ≤ b, where m is the absolute minimum and M is the absolute maximum of f on the interval[a, b], then

Use this property to estimate the value of the integral.

(smaller value)

(larger value)

m(b − a) ≤ ≤ M(b − a).f(x) dxb

3 dx16

Express the limit as a definite integral.

lim n→∞

81 + (i/n)2

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

g(s) = (t − t8)3dts

f(t) = (t − t8)3 and g(s) = , so by FTC1, g'(s) = f(s) = (s − s8)3.(t − t8)3dts

g'(r) =

g(r) = dxr

x2 + 35

G'(x) =

G(x) = cos dt6

y = dttan x

3t + t1

y = du2 u3

1 + u22 − 3x

Evaluate the integral.

(u + 3)(u − 4) du1

= = u3 − u2 − 12u = − − 12 − 0 = − (u + 3)(u − 4) du1

0(u2 − u − 12) du1

dx16 x − 3

3 sec2 t dtπ/4

where f(x) = f(x) dxπ

5 sin x if 0 ≤ x < π/25 cos x if π/2 ≤ x ≤ π

What is wrong with the equation?

is not continuous on the interval [−3, 4], so FTC2 cannot be applied. In fact, f has an infinite discontinuity at

does not exist.

x−3 dx = = 4 x−2

7288−3

The lower limit is less than 0, so FTC2 cannot be applied.

There is nothing wrong with the equation.

f(x) = x−3 is not continuous on the interval [−3, 4] so FTC2 cannot be applied.

f(x) = x−3 is continuous on the interval [−3, 4] so FTC2 cannot be applied.

f(x) = x−3 is not continuous at x = −3, so FTC2 cannot be applied.

f(x) = x−3

x = 0, so x−3 dx4

Find the derivative of the function.

g(x) = dt3x2 1

5 + t3tan x

g'(x) =

State whether the following is true or false by differentiation.

cos2x dx = x + sin 2x + C12

x + sin 2x + C = + cos 2x · 2 + 0 = + cos 2x

= + 2 cos2 x − 1 = + cos2 x − = cos2 x

Find the general indefinite integral. (Use C for the constant of integration.)

(5x2 + 6x−2) dx

dx7x3 − 8x

(7θ − 4 csc θ cot θ) dθ

sec t (5 sec t + 9 tan t) dt

8(1 + tan2 α) dα

(7 sin θ − 17 cos θ) dθπ

dx16 13

The area of the region that lies to the right of the yaxis and to the left of the parabola (the shaded region in

the figure) is given by the integral (Turn your head clockwise and think of the region as lying below the

curve from y = 0 to y = 8.) Find the area of the region.

x = 8y − y2

(8y − y2) dy.8

0x = 8y − y2

The boundaries of the shaded region are the yaxis, the line and the curve Find the area of this region

by writing x as a function of y and integrating with respect to y.

y = 8, y = 8 .x4

The velocity function (in meters per second) is given for a particle moving along a line.

(a) Find the displacement. m

(b) Find the distance traveled by the particle during the given time interval. m

v(t) = 3t − 8, 0 ≤ t ≤ 3

45. Question Details SCalc7 4.5.003.MI.SA. [2631984]

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receiveany points for the skipped part, and you will not be able to come back to the skipped part.

Tutorial Exercise

Evaluate the integral by making the given substitution.

x2 dx, u = x3 + 4x3 + 4

Tutorial Exercise

Evaluate the integral by making the given substitution.

dx, u = 1/x2sec2(1/x2)x3

Tutorial Exercise

Evaluate the indefinite integral.

x4 sin(x5) dx

Tutorial Exercise

Evaluate the indefinite integral.

dxcos xsin7 x

Evaluate the indefinite integral. (Use C for the constant of integration.)

dtcos2 t 6 + tan t3

Evaluate the definite integral.

csc 7πt cot 7πt dt1/14

If f is continuous and find

f(x) dx = 8,36

0xf(x2) dx.6

Name (AID): E5 (6607623)Submissions Allowed: 2Category: ExamCode:Locked: YesAuthor: Frith, Russell ( afrgf@uaa.alaska.edu )Last Saved: Nov 21, 2014 01:51 PM AKSTPermission: ProtectedRandomization: PersonWhich graded: Last

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Assignment Details

If a and b are positive numbers, show that

Let and du = ? .

Use this substitution to rewrite the integral in terms of u.

Then replacing u with x results in the integral

= .xa(1 − x)bdx1

0xb(1 − x)adx1

u = 1 − x. Then x =

= xa(1 − x)b dx1

0 1 ub(−du) = ub(1 − u)adu

(1 − x)adx.1

u = 1 − x. Then x = 1 − u and dx = −du, so

= = = xa(1 − x)bdx1

0(1 − u)aub(−du)0

1ub(1 − u)a du1

0xb(1 − x)adx.1

Wed Dec 3 2014 11:59 PM AKST - University of Alaska...

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