Math 113 HW #11 Solutions

[Pages:7]Math 113 HW #11 Solutions

?5.1

4. (a) Estimate the area under the graph of f (x) = x from x = 0 to x = 4 using four approx-

imating rectangles and right endpoints. Sketch the graph and the rectangles. Is your estimate an underestimate or an overestimate? Answer: Since [0, 4] has length 4, each of the four rectangles will have width 4/4 = 1, so the right endpoints are 1, 2, 3 and 4. Thus, the heights of the four rectangles are

f (1) = 1 = 1

f (2) = 2 1.414

f (3) = 3 1.732

f (4) = 4 = 2.

2

1.5

1

0.5

0

0.4

0.8

1.2

1.6

2

2.4

2.8

3.2

3.6

4

Since each rectangle has width 1, the area of the first rectangle is 1 ? 1 = 1, the area of the second is 2 ? 1 = 2, etc. Thus, we can estimate the area under the curve as

1 + 2 + 3 + 2 6.146. Since f (x) is an increasing function, this is an over-estimate of the actual area. (b) Repeat part (a) using left endpoints.

1

Answer: The endpoints of the four sub-intervals are the same, though now we're in-

terested in the left endpoints, which are 0, 1, 2, and 3. Thus, the heights of the four

rectangles are

f (0) = 0 = 0

f (1) = 1 = 1

f (2) = 2 1.414

f (3) = 3 1.732.

2

1.5

1

0.5

0

0.4

0.8

1.2

1.6

2

2.4

2.8

3.2

3.6

4

Thus, the area contained in these rectangles is

0 + 1 + 2 + 3 4.146,

which is an underestimate of the actual area.

18. Use Definition 2 to find an expression for the area under the graph of

ln x f (x) = , 3 x 10

x as a limit. Do not evaluate the limit.

Answer: Since [3, 10] has length 10 - 3 = 7, if we break this interval up into n subintervals

of equal width, each will have width x = 7/n. Then the area under the graph will be given

by

lim

n

n i=1

f (xi )x

=

lim

n

n i=1

ln xi xi

7 n

2

for any choice of sample points xi , where xi is in the ith subinterval. Choosing, say, the right endpoint of each as the sample point, we can see that

xi

=

3

+

i

7 n

,

so the above limit becomes

lim

n

n i=1

ln

3

+

i

7 n

3

+

i

7 n

7 .

n

?5.2

18. Express the limit

lim n cos xi x

n i=1

xi

as a definite integral on [, 2].

Answer: This is simply the definition of the definite integral

2 cos x dx.

x

22. Use the form of the definition of the integral given in Theorem 4 to evaluate the integral

4

(x2 + 2x - 5) dx.

1

Answer: Breaking the interval [1, 4] into n subintervals of equal width, each will be of width

4-1 3

x =

=.

nn

Moreover, the right endpoint of the ith subinterval will be

3

xi

=

1

+

i. n

Therefore, the height of the ith rectangle will be (since we're using right endpoints),

f (xi) = x2i + 2xi - 5

32

3

= 1+i +2 1+i -5

n

n

=

1

+

6 i n

+

i2

9 n2

+

2

+

6 i n

-

5

=

i2

9 n2

+

12 i

n

-

2.

3

Therefore,

4

n

1

(x2 + 2x - 5) dx

=

lim

n

f (xi)x

i=1

n

= limn

i2

9 n2

+

12 i

n

-

2

3 n

i=1

n

= lim

n

i2

27 n3

+

36 i n2

-

6 n

i=1

= lim

n

n

i2

27 n3

+

n

36 i n2 -

n

6 n

i=1

i=1

i=1

= lim

n

27 n3

n

i2

+

36 n2

n

6

i-

n

n

1

i=1

i=1

i=1

Therefore, since

n

1=1

i=1

n

n(n + 1)

i=

2

i=1

n

i2

=

n(n

+

1)(2n

+

1) ,

6

i=1

we see that the above limit is equal to

27 n(n + 1)(2n + 1) 36 n(n + 1) 6

54n3 + 81n2 + 27n 36n2 + 36n

lim

n

n3

6

+ n2

2

- n = lim

n

n

6n3

+ 2n2 - 6

54 36 = + -6

62

= 9 + 18 - 6

= 21.

Therefore,

4

(x2 + 2x - 5) dx = 21.

1

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

(a)

2 0

g(x)dx

Answer: Since on [0, 2] the graph of g(x) is just a straight line of slope -2 coming down

from y = 4 to y = 0, the area is just the area of the triangle

1 2 ? 4 = 4.

2

Since this area is above the x-axis, definite integral equals the area, so

2 0

g(x)dx

=

4.

4

(b)

6 2

g(x)dx

Answer: On [2, 6] the graph of g(x) is a semi-circle of radius 2 lying below the x-axis.

Its area is

1 (2)2 = 2. 2

Since it lies below the axis, the integral is negative, so

6

g(x)dx = -2.

2

(c)

7 0

g(x)dx

Answer: Since

7

2

6

7

7

g(x)dx = g(x)dx + g(x)dx + g(x)dx = 4 - 2 + g(x)dx,

0

0

2

6

6

we just need to determine

7 6

g(x)dx.

Since this is a straight line of slope 1 going up

from the x-axis (at x = 6) to y = 1 (at x = 7), it describes a triangle of area

1

1

1?1= .

2

2

Since this area lies above the axis,

7 6

g(x)dx

=

1/2,

so

7

7

19

g(x)dx = 4 - 2 + g(x)dx = 4 - 2 + = - 2 -1.78.

0

6

22

44. Use the result of Example 3 to compute

3

(2ex - 1)dx.

1

Answer: Example 3 says that

3 1

exdx

=

e3

-

e,

we

need

to

use

the

properties

of

the

definite

integral to express the given integral in terms of

3 1

ex

dx.

Now, by Property 4,

3

3

3

(2ex - 1)dx = 2ex - 1dx.

1

1

1

In turn, by Property 1,

3

1dx = 1(3 - 1) = 2.

1

By Property 3,

3

3

2exdx = 2 exdx.

1

1

Putting these together, then,

3

3

(2ex - 1)dx = 2 exdx - 2.

1

1

Plugging in the value we know for

3 1

exdx,

we

see

that

3

(2ex - 1)dx = 2(e3 - e) - 2 = 2(e3 - e - 1) 32.73.

1

5

?5.3

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

x2

h(x) =

1 + r3 dr.

0

Answer: Make the change of variables u = x2. Then

d h (x) =

dx

x2

1 + r3 dr

d =

0

dx

u

1 + r3 dr .

0

By the Chain Rule, this is equal to

d

u

1 + r3 dr

du .

du 0

dx

Using

the

Fundamental

Theorem

and

the

fact

that

du dx

=

2x,

we

see

that

h (x) = 1 + u3 (2x) = 1 + (x2)3 (2x) = 2x 1 + x6.

26. Evaluate the integral

2

cos d.

Answer: Since sin is an antiderivative of cos , the second part of the Fundamental Theorem

says that

2

2

cos d = sin = sin 2 - sin = 0 - 0 = 0.

36. Evaluate the integral

1

10x dx.

0

Answer: Since

d (10x) = 10x ln 10, dx

we see that

10x

ln 10

is an antiderivative of 10x. Therefore,

1

10xdx =

10x 1 10

=

-

1

=

9 .

0

ln 10 0 ln 10 ln 10 ln 10

6

40. Evaluate the integral Answer: Re-write the integral as

2 4 + u2 1 u3 du.

2 1

4 u2 u3 + u3

2

2

du = 4u-3du + u-1du.

1

1

Then,

since

u-2 -2

=

-

1 2u2

is

an

antiderivative

for u-3

and

since

ln u

is

an

antiderivative for

u-1, we see that the above is equal to

-1 2

2

22

3

4 2u2

+

1

ln u =

1

-+ 41

+ (ln 2 - ln 1) = + ln 2. 2

7

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