Can someone please check the answer that I got please....
I don't think it can be done is that correct
is that right
@phi is that answer correct
You should know part of the answer... the integral of b(x) but do you know how to integrate \[ \int_4^9 x \ dx\] ??
@electrokid can you please help me
when you get time, watch http://www.khanacademy.org/math/calculus/integral-calculus/definite_integrals/v/evaluating-simple-definite-integral and maybe some of the videos in the intro to calculus.
**Can someone please check the answer that I got please.... what answer would that be ?
I posted it above Cannot be done
yes, you can do this one, but you have to integrate x dx from 4 to 9 \[ 2\int_4^9 x \ dx = ?\]
10
see http://www.khanacademy.org/math/calculus/integral-calculus/indefinite_integrals/v/indefinite-integrals-of-x-raised-to-a-power for how to integrate this
is 37 the answer
is that the correct answer
can you show your work ?
yup 2(10)+17=37
yup 2(10)+17=37
the 17 is correct but how did you get the 10 ? you have to integrate x dx can you do that ?
You have to solve \[ 2 \int_4^9 x \ dx \] see http://www.khanacademy.org/math/calculus/integral-calculus/indefinite_integrals/v/indefinite-integrals-of-x-raised-to-a-power for how to do this.
I got 10 from 2(9)-2(4)=10
Yes, but that is not how to do an integral. Erase that from your memory. use the rule \[\int x^n dx = \frac{x^{n+1}}{n+1} \] in your problem n is 1
so then is the overall answer 19
what is the integral? It is: \[ \int x^1 dx = \frac{x^{1+1}}{1+1} = \frac{x^2}{2}\] we want \[ 2\ \int x dx = 2\cdot \frac{x^2}{2}= x^2\] however we have lower limit of 4 and an upper limit of 9. that means you evaluate x^2 at x=9 and subtract x^2 at x=4 can you do that ?
final answer 65
yes \[ 2 \int_4^9 x \ dx = 65\] now add in the integral of b(x) = 17 for the final answer.
final answer 82 correct?
yes. But if I were you, I would learn how to integrate these simple functions. see Khan's videos.
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