graphing and evaluating definite integrals

[chillintoucan28]

the integral of: from x=0 to x=4 of x+2, from x=0 to x=5 of square root of (25-x^2), from x=0 to x=4 of 2-x, from x=0 to x=4 of absolute value of x-2.

Also, since I need to graph these, can you tell me the best method I should use to find the area under the graph. Which of these methods is best: the start point, mid point, end point, or trapezoid rule?

Also for each of these could you show me how used solved it algebraically, you know if you used the fundamental theorem of calculus?

i know this is alot, but it would be greatly appreciated. Thanks alot!

 

[Deleted member 4993]

the integral of: from x=0 to x=4 of x+2, from x=0 to x=5 of square root of (25-x^2), from x=0 to x=4 of 2-x, from x=0 to x=4 of absolute value of x-2.

Also, since I need to graph these, can you tell me the best method I should use to find the area under the graph. Which of these methods is best: the start point, mid point, end point, or trapezoid rule?

I am assuming you are going to use "numerical" methods - why don't you use a spreadsheet (like excel) and find out the error for each method? Your textbook should provide you with formulae for bounding such errors.

Also for each of these could you show me how used solved it algebraically, you know if you used the fundamental theorem of calculus?

Are you saying you do not know how to integrate these functions? Have you taken calculus at all?

i know this is alot, but it would be greatly appreciated. Thanks alot!

duplicate post:

http://answers.yahoo.com/question/index ... 400AAL4Hyq

Please share with us your work/thoughts - so that we know where to begin to help you.

 

[skeeter]

the integral of:

every one of these integrals can be evaluated just by sketching a diagram and using simple geometric formulas for area ...

from x=0 to x=4 of x+2 a trapezoid

from x=0 to x=5 of square root of (25-x^2) a very "nice" sector of a circle

from x=0 to x=4 of 2-x two triangles ... one above the x-axis, one below ... something "nice" about their respective areas that makes the integral evaluation incredibly simple

from x=0 to x=4 of absolute value of x-2 two triangles again ... only this time they're both above the x-axis