Question

Evaluate the integral.
http://www.webassign.net/cgi-bin/symimage.cgi?expr=int_%28-2%29%5E%281%29%20%5C%28x%5E3-6%20x%5C%29%20dx

Answer 1

im lost pleas help

Answer 2

What have you tried?

Answer 3

i got the answer to be 3/4 but its wrong..i did the anti dertivitve..but dont know

Answer 4

The antiderivative is\[\frac{x^4}{4}-3x^2.\]Was that your result?

Answer 5

no...is that the answer? cuz i did it differently..

Answer 6

its 6x^2/6

Answer 7

That's not right. You must find the antiderivative of \[x^3\]and then the antiderivative of\[-6x\]and add them. The general rule is that the antiderivative of \[x^n\]is\[\frac{x^{n+1}}{n+1}\](for \[n\neq-1).\]For\[x^3\]you have n = 3 so the antiderivative is\[\frac{x^{3+1}}{3+1}=\frac{x^4}{4}.\]Try to find the antiderivative of -6x by yourself.

Answer 8

-6^2/7?

Answer 9

\[-6x^2/7\]

Answer 10

You have to add one to the exponent. Forget the -6 in front and try to find the antiderivative of\[x^2,\]and then you multiply the result by -6. I'll give you another example: finding the antiderivative of\[x^6:\]You add one to the exponent and divide by the exponent plust one. Because the exponent is 6, it becomes 7 (we add one) and we divide by 7 (the exponent plus one), so it becomes\[\frac{x^7}{7}.\]

Answer 11

o wait i meant -6x^2/3 which becomes -2x^2?

Answer 12

You need to change the exponent too. It's \[-6 \frac{x^3}{3} = -2x^3.\]

Answer 13

o alright so next

Answer 14

Think about it this way rmalik. If you have \[\int\limits x^2dx=\frac{x^3}{3}+C\].
So you want to know if this is correct; differentiate it.
You have:
\[\frac{d}{dx}(\frac{x^3}{3}+C)=3(\frac{x^2}{3})=x^2\]
Which is what you started with. The -6 (or any non-zero constant) can be pulled out because we know that:
\[\int\limits C*f(x)dx=C*\int\limits f(x)dx\]

Answer 15

Oh, sorry it was -6x originally so it becomes \[-6 \frac{x^2}{2} = -3x^2\](add one to the exponent and divide by the exponent plus one).

Answer 16

Now we have the antiderivative\[F(x) = \frac{x^4}{4} - 3x^2.\]The value of the integral is\[\int_{-2}^1 (x^3 - 6x) dx=F(1) - F(-2)\]where F is the antiderivative. You just have to plug 1 and -2 into the formula.

Answer 17

so its -10.75?

Answer 18

http://www.wolframalpha.com/input/?i=integral+of+x^3-6x%2Cx%2C-2%2C1

Answer 19

\[F(1) = \frac{1^4}{4} - 3 \cdot 1^2 = \frac{1}{4} - 3 = -\frac{11}{4}\]\[F(-2) = \frac{(-2)^4}{4} - 3 \cdot (-2)^2 = \frac{16}{4} - 3 \cdot 4 = 4 - 12 = -8\]so\[F(1) - F(-2) = -\frac{11}{4} - (-8) = 8 - \frac{11}{4} = \frac{21}{4}.\]

Answer 20

ooo darn i did it other way..i didnt combine it...well now i see what i made..thanks

Answer 21

If you need more help malik just keep posting :P

Answer 22

ok thanks malevolence19 : )

Answer 23

No problem xD