1) What is the length of the graph of y = x 2 from x = -1 to x = 1? 2) A package

Question

1) What is the length of the graph of y = x2 from x = -1 to x = 1?

2) A package of frozen strawberries has been taken from the freezer which is at -25 degrees Fahrenheit and placed on the counter in a room in which the temperature is 78 degrees. Thirty minutes later, the temperature has risen to 0 degrees. The rate at which an object changes temperature is directly proportional to the difference in the temperature of the object and the temperature of the surrounding medium.

a) Give a formula that gives the temperature of the berriees in terms of time.

b) When will the berries reach 32 degrees?

3) a) Estimate the value of 01 sin2 (x)dx using Simpson's rule and n = 4

b) The error formula for Simpson's rule is given by ((b-a)5K4)/180n4 where K4 is an upper bound on the absolute value of the fourth derivative of f(x) on [a,b]. How accurate is the answer in part a?

4) Find the integrals

a) xe2xdx

b) dx/(4+x2)1/2

c) sin(ln(x))dx

d) (1-cos2()d)1/2 from 0 to

e) sin3xdx

f) dx/(x2+2x+5)

g) dx/(x3+x)

h) xdx/(x2+3x+2)

Explanation / Answer

y = x2 from x = -1 to x = 1

Length L = integral ds

ds = sqrt(1 +(dy/dx)2) dx , -1 <=x <= 1

dy/dx = 2x2-1 = 2x

==> L = integral [-1 to 1] sqrt(1 +(dy/dx)2) dx

==> L = integral [-1 to 1] sqrt(1 +(2x)2) dx

==> L = integral [-1 to 1] sqrt(1 + 4x2) dx

==> L = integral [-1 to 1] sqrt(4x2 +1) dx

put x = (tanu)/2 == u = tan-1(2x)

==> dx = (sec2u)/2 du

==> L = integral [-1 to 1] sqrt(4((tanu)/2 )2 +1) (sec2u)/2 du

==> L = integral [-1 to 1] sqrt(tan2u +1) (sec2u)/2 du            since sec2x - tan2x = 1 ==> sec2x = 1 + tan2x

==> L = (1/2) integral [-1 to 1] sqrt(sec2u) sec2u du

==> L = (1/2) integral [-1 to 1] sec3u du

==> L = (1/2) integral [-1 to 1] secu sec2u du

consider integral sec3u du = integral secu sec2u du

we know that integral UV' = UV - integral U'V     (integration by parts)

U = secu ==> U' = secu tanu, V' = sec2u ==> V = tanu

==> integral sec3u du = {secu tanu - integral secu tanu tan u du }

==> integral sec3u du = {secu tanu - integral secu tan2u du }

==> integral sec3u du = {secu tanu - integral secu (sec2u -1) du }

==> integral sec3u du = {secu tanu - integral sec3u - secu du }

==> integral sec3u du = {secu tanu - integral sec3u du + integral secu du }

==> integral sec3u du + integral sec3u du = {secu tanu + integral secu du}

==> 2 integral sec3u du = secu tanu + ln(secu + tanu )

==> integral sec3u du = (1/2) {secu tanu + ln(secu + tanu )}

==> L = (1/2) integral [-1 to 1] sec3u du = (1/2) [-1 to 1] (1/2) [secu tanu + ln(secu + tanu )]

==> L = (1/4) [-1 to 1] {sec(tan-1(2x)) tan(tan-1(2x)) + ln(sec(tan-1(2x)) + tan(tan-1(2x))) - [sec(tan-1(2x)) tan(tan-1(2x)) + ln(sec(tan-1(2x)) + tan(tan-1(2x))]}

L = (1/4) {sec(tan-1(2)) tan(tan-1(2)) + ln(sec(tan-1(2)) + tan(tan-1(2))) - [sec(tan-1(-2)) tan(tan-1(-2)) + ln(sec(tan-1(-2)) + tan(tan-1(-2))]}

==> L = (1/4) {5.91578 - [-5.91578]}

==> L = (1/2){5.91578}

==> L = 2.9579

There length of graph from -1 to 1 = 2.9579 units

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