Calculus Problems and Solutions

Calculus Problems and Solutions

[1]Evaluate 3xdx

= Let the integrating factor  be In

In =

Let U be   and dV =

=du=2cos x sin2xdx , v = -cos x.

Hence =  +

[2]Evaluate3x. cos2xdx

Let cosx be U. then we have

=du = -sinx dx

=dx =

Hence =

=

=cos2x + C

[3] Evalaute

=

=  =

[4]Evaluate

=dz

=(x2z +2z2 -5z)-1

[1] Approximate   using (a) left endpoints, (b) right endpoints (c) midpoints and n= 1000 partitioning intervals. Explain which technology  you used in the estimate. Fill out the table

Let’s get the interval of the points. That is

b=1 and a =0 n=1000

Then     = 0.001

The interval is 0.001

And the width of each rectangle formed will be      = 0.25

= =

Y =

Left points = choose xi* =xi

R4 =f(xi*)

   = 0 0.001 +  = 0.0001579.

Right points =[f(xi+1) f(xi+2)+ f(xi+3)

=0 +  = 0.001887

Midpoint = reftp[oints +right points then you divide by 2.

= 0.0010225

Error =

=where m =| f’’(x)|

=  = 1.3706 * 107

Type Approximation errror
Left

Right

Midpoint

.

0.001887

0.0010225

1.3706 * 107

[2]Let f(x) =. Sketch an approximate graph of F on the interval [ 0,  ] by filling out this table and then plotting the corresponding points.

Interval =  = 0.01772.

Y =[cost2

T = F(x)
0

0.5

1

1.5

=

0

0.56

0.8952

0.99952

1.99

Fill the table using n =100 intervals. Explain what technology you used. I Integrated the equation using normal formula with respect to dt. I went ahead to get the point intervals which helped me to get the widths.

[3] Use the error estimation formula to determine how many partitioning intervals is needed to ensure that the midpoint rule approximations for = are accurate within 0.01

Error =

=

= but we get m from | F’’(x)|

M= 0.000048 after integration and the limits be 2 and 3.. Hence the error =  0.02.

[4] (a) Graph y=

Y2 = 1-x2 the table below shows the points to plot

Y x
1 0
0 1

(b) Partitioning into how many intervals does insure that  can be approximated using the midpoint rule to within 0.01?

( c) Then use the technology  to compute an approximation of A =  accurate to within 0.01

Get the left points = R4 =f(xi*)

 = then

1* 0.01+ 1*0.001 +1*0.001 +0 = 0.003

Right points = [f (xi+1) f(xi+2)+ f(xi+3)

   = by integration method the right point is = 0.4563

Hence the midpoint = left + right and then divide by 2.

0.003+  0.4563 = 0.22965

A =0.22965

(d) Finally calculate the number  P= ( A- ). What number does 6p approximate?

P =( A- ). Where by A = 0.22965.then we substitute

=(0.22965 ) = 0.099441366

P = 0.099441366

If p =0.099441366 what of 6p?

=0.59664

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