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I need to find the definite integral between 3 and -2 (i don't know how to make the integral sign on here) (x^4 - 9x^3 + 7x^2 - 3x +1)dx

my friends and i have tryed this a bunch of times and for some reason we keep getting different answers. So please help?!

2006-11-01 14:34:58 · 5 answers · asked by iamthehandcuff 1 in Science & Mathematics Mathematics

5 answers

Integrating that function gives you:

(x^5)/5 - (9x^4)/4 + (7x^3)/3 - (3x^2)/2 + x

To take the defiunite integral from -2 to 3 you plug those both in and subtract, that gives:

((243/5) - (729/4) + (189/3) - (27/2) + 3) - ((-32/5) - (144/4) + (-56/3) - (12/2) - 2)

Doing the arithmetic gives: -12.08333333

If you take the integral from 3 to -2 you get the opposite: 12.083333333

2006-11-01 14:45:08 · answer #1 · answered by Nobody 3 · 0 0

Use power rule on each term to get:

(1/5)x^5 -(9/4)x^4 + (7/3)x^3 -(3/2)x^2 + x

Now evaluate the above expression first with x = 3 and then with x = -2.

Next subtract the value obtained with x = -2 from the value obtained with x=3 for your final answer.

With x = 3 you should get:
(1/5)(243)- (9/4)(81) + (7/3)(27) - (3/2)(9) +3

With x = -2 you should get:
(1/5)(-32)- (9/4)(16) + (7/3)(-8) - (3/2)(4) -2

It should be simple to use your calculator to get the above values and then divide them to get final answer.

2006-11-01 14:57:23 · answer #2 · answered by ironduke8159 7 · 0 0

I am not sure but I think you use the power rule for integration......it is: x^n dx = (1/ n+1)x^n+1 + C (n cannot = -1)..... so in words,
the derivivate of x raised to the n power is equal to 1 over n plus 1 all multiplied by x raised to the n plus one power plus a constant....hope this helps

2006-11-01 14:45:28 · answer #3 · answered by chalupa1769 2 · 0 0

you could no longer combine it indefinitely to get an uncomplicated applications, yet you could combine it between -infinity and infinity: because of the fact the function is even, evaluate int{from 0 to infinity}[(sin(x))/x]dx. combine by skill of components, utilising u = a million/x, dv = sin(x)dx to get int{from 0 to infinity}[(sin(x))/x]dx = int{from 0 to infinity}[[a million - cos(x)]/[x^2]]dx. permit a million/x^2 = int{from 0 to infinity}[te^(-tx)dt]. placed that interior the fundamental to make a double fundamental. Now swap the order of integration interior the double fundamental (justified for nonnegative applications), and compute the fundamental.

2016-11-26 22:59:34 · answer #4 · answered by Anonymous · 0 0

I got -12.083 if the 3 is on top (+12.083 if the -2 is on top)

2006-11-01 14:52:07 · answer #5 · answered by Steve 7 · 0 0

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