Integral approach help

[Imum Coeli]

I have two separate questions, both of which I am not sure if I am approaching with the right method.

1) Question:
Find an antiderivative of g(t) = t*sqrt(t^2 +9). Hence calculate the integral from 0 to 4 of g(t) dt.

Notes:
I have a solution to this but I found it by trial and error and some guessing. This is not a very satisfactory method and I was wondering what, if any, the correct method and notation would be?

2) Question:
Let F(x) = integral from root x to 2 of ln(t) dt where x > 0. Find F''(x) = d^2F(x)/dx^2.

Notes:
I tried solving the integral and then finding the second derivative but it got kind of messy and I think that there must be a better way to do it. So I was wondering if there is something I'm not seeing?

If anyone could point me in the right direction I would be very grateful.

 

[tkhunny]

1) "Correct"? What a funny concept. Unique answers don't care how you find them. Of course, some methods have more or less pain than others.Substitution. \(\displaystyle u = t^{2} + 9\)

2) Your idea should work, and it's not really that hard, but there are easier ways. Here's one.

You have g(t). It has an antiderivative G(t). Then \(\displaystyle \int g(x)\;dx\;=\;G(x) + C\). You should believe this.

If, then, we have \(\displaystyle F(x) = \int\limits_{a}^{b}g(t)\;dt\), we should have \(\displaystyle F(x) = G(t)|_{a}^{b} = G(b) - G(a)\)

Are you following? We're about to extend it.

If, then, we have \(\displaystyle F(x) = \int\limits_{x}^{2}g(t)\;dt\), we should have \(\displaystyle F(x) = G(t)|_{x}^{2} = G(2) - G(x)\)

Finally, we have \(\displaystyle F'(x) = (d/dx)G(2) - (d/dx)G(x) = 0 - G'(x) = -g(x)\)

That little g on the end is the same g that was in the integral! How does this help you?

 

[Imum Coeli]

That's what I have thought previously. An answer by hook or by crook...

1) That's much nicer. Thanks.

2) So then I just take the derivative of -g(x)? (which is again much nicer).


Is there any way to show explicitly that F''(x)=d^2F(x)/dx^2?
For example, is something like the below 'correct'. I think I'm slightly confused about the notation...

F'(x) = 0 - (d/dx)G(x)
F''(x) = 0 - (d/dx)((d/dx)G(x)) = 0 - (d^2/dx^2)G(x)


Thanks for your help.

 

[Imum Coeli]

So using x^(1/2) we have something like...

F(x) = integral from x^(1/2) to 2 of ln(t) dt.
F(x) = F(2) - F(x^(1/2))

taking the derivative
F'(x) = d/dx (F(2) - F(x^(1/2)))
F'(x) = 0 - f(x^(1/2))*1/2*x^-(1/2)
F'(x) = -1/(2*sqrt(x))*ln(x^(1/2))
F'(x) = -1/(4*sqrt(x))*ln(x)

taking the second derivative
F''(x) = d/dx (-1/(4*sqrt(x))*ln(x))
F''(x) = (sqrt(x)*ln(x)-2*sqrt(x))/(8*x^2)

Does that show that F''(x) = d^2F(x)/dx^2 ?

 

[DrPhil]

2) Question:
Let F(x) = integral from root x to 2 of ln(t) dt where x > 0. Find F''(x) = d^2F(x)/dx^2.

Notes:
I tried solving the integral and then finding the second derivative but it got kind of messy and I think that there must be a better way to do it. So I was wondering if there is something I'm not seeing?

If anyone could point me in the right direction I would be very grateful.

Is there any way to show explicitly that F''(x)=d^2F(x)/dx^2?
For example, is something like the below 'correct'. I think I'm slightly confused about the notation...

The statement F''(x)=d^2F(x)/dx^2 IS the notation .. you don't have to "prove" that.

\(\displaystyle \displaystyle F(x) = \int_{\sqrt{x}}^2 \ln{t}\ dt \)

The lower limit is a function of x, but not simply "x".
Let \(\displaystyle u = \sqrt{x}, \;\;\;\;\;\frac{du}{dx} = \frac{1}{2 \sqrt{x}} \)

By the Fundamental Theorem of Calculus, the derivative of F with respect to the lower limit, u, is the negative of the value of the integrand at the lower limit:

\(\displaystyle \displaystyle \dfrac{dF}{du} = - \ln{\sqrt{x}}\)

\(\displaystyle \displaystyle F'(x) = \dfrac{dF}{dx} = \dfrac{dF}{du}\dfrac{du}{dx}\)

.....................\(\displaystyle \displaystyle=- \ln{\sqrt{x}} \times \dfrac{1}{2 \sqrt{x}} \)

That is the same as you got. I also agree with your differentiation to get F''.

\(\displaystyle \displaystyle F''(x) = \dfrac{(\ln{x} - 2) \sqrt{x}}{8 x^2} \)

 

[Imum Coeli]

Cool. Thanks.