1. Adding Integrals and Velocity Function

Hello All-
I was having some difficulty with the below problem.

from 0 to 3 [f(x)] dx = 6
from 3 to 5 [f(x)] dx = 4

Find the from 0 to 5 [ 3 + 2f(x)] dx

My thought was that the answer would simply be 23. [3+2(4+6)]=23
But this seems too simple. Does anyone know?

Thanks!

Also, one more problem I was having trouble with:

v(t) = (t^2)[ln(t+2)]
Find the acceleration at t=6

All you would have to do I think would be to take the derivative of the velocity function and plug in 6 for t, but I'm not positive about this, and even if this was right, I'm not sure how to take the derivative of this particular function.

Any help is appreciated!!

2. Originally Posted by pandaexpress915
Hello All-
I was having some difficulty with the below problem.

from 0 to 3 [f(x)] dx = 6
from 3 to 5 [f(x)] dx = 4

Find the from 0 to 5 [ 3 + 2f(x)] dx

My thought was that the answer would simply be 23. [3+2(4+6)]=23
But this seems too simple. Does anyone know?

Thanks!

Also, one more problem I was having trouble with:

v(t) = (t^2)[ln(t+2)]
Find the acceleration at t=6

All you would have to do I think would be to take the derivative of the velocity function and plug in 6 for t, but I'm not positive about this, and even if this was right, I'm not sure how to take the derivative of this particular function.

Any help is appreciated!!
Dear pandaexpress915,

$\displaystyle \int_{0}^{5}{\left(3+2f(x)\right)}dx=\int_{0}^{5}{ 3}dx+2\int_{0}^{5}{f(x)}dx=3\int_{0}^{5}dx+2\left( \int_{0}^{3}{f(x)}dx+\int_{3}^{5}{f(x)}dx\right)$

Hope you can continue from here....

For your second question, you have to use the product rule and the chain rule of differentiation. Please refer,

Product rule: Product rule - Wikipedia, the free encyclopedia

Chain rule: Chain rule - Wikipedia, the free encyclopedia

3. Originally Posted by pandaexpress915
Hello All-
I was having some difficulty with the below problem.

from 0 to 3 [f(x)] dx = 6
from 3 to 5 [f(x)] dx = 4

Find the from 0 to 5 [ 3 + 2f(x)] dx

My thought was that the answer would simply be 23. [3+2(4+6)]=23
But this seems too simple. Does anyone know?
Almost that simple! $\displaystyle 1\int_2^5 f(x) dx= 2(10)= 20$, of course, but $\displaystyle \int_2^5 2dx= 3(5- 2)= 9$.

Thanks!

Also, one more problem I was having trouble with:

v(t) = (t^2)[ln(t+2)]
Find the acceleration at t=6

All you would have to do I think would be to take the derivative of the velocity function and plug in 6 for t, but I'm not positive about this, and even if this was right, I'm not sure how to take the derivative of this particular function.

Any help is appreciated!!
Yes, that's exactly what you do. And you take the derivative by using the product rule: $\displaystyle (t^2)' ln(t+2)+ t^2(ln(t+2))'$
I presume you know that the derivative of $\displaystyle t^2$ is 2t. And the derivative of ln(t+2) is $\displaystyle \frac{1}{t+2}$

4. Originally Posted by HallsofIvy
Almost that simple! $\displaystyle {\color{red}1\int_2^5 f(x) dx= 2(10)= 20}$, of course, but $\displaystyle \int_2^5 2dx= 3(5- 2)= 9$.

Yes, that's exactly what you do. And you take the derivative by using the product rule: $\displaystyle (t^2)' ln(t+2)+ t^2(ln(t+2))'$
I presume you know that the derivative of $\displaystyle t^2$ is 2t. And the derivative of ln(t+2) is $\displaystyle \frac{1}{t+2}$
Dear HallsofIvy,

Can you please tell me how you got $\displaystyle 1\int_2^5 f(x) dx= 2(10)= 20$ ??

5. A mistype on my part- of course, I meant that $\displaystyle 2\int_0^5 f(x)dx= 2(10)= 20$ and [tex]\int_0^5 3dx= 3(5- 0)= 15[/quote]. Thanks for catching that.