**Calculus Problems and Solutions**

[1]Evaluate ^{3}xdx

= Let the integrating factor be I_{n}

I_{n} =

Let U be and dV =

=du=2cos x sin^{2}xdx , v = -cos x.

Hence = +

[2]Evaluate^{3}x. cos^{2}xdx

Let cosx be U. then we have

=du = -sinx dx

=dx =

Hence =

=

=cos^{2}x + C

[3] Evalaute

=

= =

[4]Evaluate

=dz

=(x^{2}z +2z^{2 }-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 x_{i}* =x_{i}

R_{4} =*f(x*_{i}**)*

* =* 0 0.001 + = 0.0001579.

Right points =[f*(x*_{i+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 * 10^{7}

Type | Approximation | errror |

Left
Right Midpoint |
.
0.001887 0.0010225 |
1.3706 * 10^{7} |

[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 =[cost^{2}

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=

Y^{2} = 1-x^{2} 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 = R_{4} =*f(x*_{i}**)*

* =** then*

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

Right points = [f *(x*_{i+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