# Thread: [SOLVED] Derivative of an integral

1. ## [SOLVED] Derivative of an integral

From galactus,
$\frac{d}{dx}\int_{h(x)}^{g(x)}f(t)dt=f(g(x))g'(x)-f(h(x))h'(x)$
. It helped me verifying my result and confirmed it.
I got that the derivative of $\int_a^{x^3} sin^3(t) dt=3x^2{sin(x^3)}^3$.
While from Mathematica, it gives the result $\frac{1}{12} \big (9cos(a)-cos(3a)-9cos(x^3)+cos(3x^3)\big)$ which I doubt is equal to my result. Can you confirm my result (if yes, I will think that the result from Mathematica is the same, even if it seems more than incredible.). Thanks!!

2. Originally Posted by arbolis
From galactus, . It helped me verifying my result and confirmed it.
I got that the derivative of $\int_a^{x^3} sin^3(t)^3 dt=3x^2{sin(x^3)}^3$.
While from Mathematica, it gives the result $\frac{1}{12} \big (9cos(a)-cos(3a)-9cos(x^3)+cos(3x^3)\big)$ which I doubt is equal to my result. Can you confirm my result (if yes, I will think that the result from Mathematica is the same, even if it seems more than incredible.). Thanks!!

3. Ok. How it is possible that Mathematica gives an answer in function of a? Maybe it doen't consider it as a constant. I don't know at all.

4. Originally Posted by arbolis
Ok. How it is possible that Mathematica gives an answer in function of a? Maybe it doen't consider it as a constant. I don't know at all.
Yeah, most likely.

5. Originally Posted by arbolis
Ok. How it is possible that Mathematica gives an answer in function of a? Maybe it doen't consider it as a constant. I don't know at all.
I don't think that you should use the formula given by galactus if any one of the limit is constant.suppose if both limits were constant then you would have got 0 as result which is not necessary.
Since you were solving definite integral getting you can not take even constant for granted as we do in indefinite integral(giving general term e.g c for all constant terms) but in definite integral showing constant term is must.

6. Originally Posted by nikhil
I don't think that you should use the formula given by galactus if any one of the limit is constant.suppose if both limits were constant then you would have got 0 as result which is not necessary.
Since you were solving definite integral getting you can not take even constant for granted as we do in indefinite integral(giving general term e.g c for all constant terms) but in definite integral showing constant term is must.
What are you talking about? The constant terms will drop out due to the derivative.

7. Originally Posted by arbolis
Ok. How it is possible that Mathematica gives an answer in function of a? Maybe it doen't consider it as a constant. I don't know at all.
One way to check this would be to let a = 0, say, and then put Mathematica on the job. Note that there are various equivalent forms of the correct answer.

8. I don't think that you should use the formula given by galactus if any one of the limit is constant.suppose if both limits were constant then you would have got 0 as result which is not necessary.
Since you were solving definite integral getting you can not take even constant for granted as we do in indefinite integral(giving general term e.g c for all constant terms) but in definite integral showing constant term is must.
I don't understand all what you mean, but the formula galactus gave works here. (Now I realize I proved it about 9 months or so ago.) And if both of the limit of the integral were constants, say 0 and 10, then the derivative would be 0, which is true. Think about it, you have $F(x)=\int_0^{10} f(t)dt$. It's obvious that the integral is equal to a constant, whatever x is, since it doesn't depends of x. So the derivative of F is 0 (since the derivative of a constant is always 0).

9. Originally Posted by nikhil
I don't think that you should use the formula given by galactus if any one of the limit is constant.suppose if both limits were constant then you would have got 0 as result which is not necessary.
Since you were solving definite integral getting you can not take even constant for granted as we do in indefinite integral(giving general term e.g c for all constant terms) but in definite integral showing constant term is must.
The formula given by galactus works fine, including the case where one or both of h(x) and g(x) are equal to a constant.

10. ## Ooops

Originally Posted by Mathstud28
What are you talking about? The constant terms will drop out due to the derivative.
oops sorry dude forgot we were talking about its derivative. Next time i will be carefull for sure

11. ## Ooops

Originally Posted by Mathstud28
What are you talking about? The constant terms will drop out due to the derivative.
oops sorry dude forgot we were talking about its derivative at the end. Next time i will be carefull for sure

12. One way to check this would be to let a = 0, say, and then put Mathematica on the job. Note that there are various equivalent forms of the correct answer.
If I typed it well, I got that the derivative equals $8 x^2 Cos(\frac{x^3}{2})(2+Cos(x^3))Sin^3(\frac{x^3}{2})-4x^2Sin^4(\frac{x^3}{2})Sin(x^3)$. It surprises me more and more. If somebody has Mathematica and has some time to spend, I'll be glad if you could check out the derivative of the integral!

13. Originally Posted by arbolis
From galactus, . It helped me verifying my result and confirmed it.
I got that the derivative of $\int_a^{x^3} sin^3(t)^3 dt=3x^2{sin(x^3)}^3$.
While from Mathematica, it gives the result $\frac{1}{12} \big (9cos(a)-cos(3a)-9cos(x^3)+cos(3x^3)\big)$ which I doubt is equal to my result. Can you confirm my result (if yes, I will think that the result from Mathematica is the same, even if it seems more than incredible.). Thanks!!
You have as the integrand $sin^3(t)^3$ .....? Is the integrand $\sin^3 (t)$ or $\sin^3(t^3)$?

14. You have as the integrand .....? Is the integrand or ?
Sorry about that, it is $\sin^3 (t)$, that is $(sin(t))^3$.

EDIT: Finally I don't mind what Mathematica thinks. Since I didn't made this program, it doesn't interest me that much. It may has some conventions I don't know yet, that's all. Thread solved.