# Thread: derivative of an integral

1. ## derivative of an integral

I'm struggling with this: $\frac{d}{dx}\int_{x}^{\pi/2} sin\frac{t}{2}cos\frac{t}{3} \, dt$

Using $\frac{d}{dx}\int_{a}^{g(x)} f(t) \, dt =f(g(x)) g'(x)$, I get $\frac{sin\frac{x}{2}sin\frac{x}{3}}{3}-\frac{cos\frac{x}{3}cos\frac{x}{2}}{2}$, which is wrong (I think).

2. Originally Posted by th%$&873 I'm struggling with this: $\frac{d}{dx}\int_{x}^{\pi/2} sin\frac{t}{2}cos\frac{t}{3} \, dt$ Using $\frac{d}{dx}\int_{a}^{g(x)} f(t) \, dt =f(g(x)) g'(x)$, I get $\frac{sin\frac{x}{2}sin\frac{x}{3}}{3}-\frac{cos\frac{x}{3}cos\frac{x}{2}}{2}$, which is wrong (I think). yes that is wrong. apply a minus sign to change the limits: $\frac{d}{dx}\int_{x}^{\pi/2} sin\frac{t}{2}cos\frac{t}{3} \, dt = - \frac{d}{dx}\int_{\pi / 2}^{x} sin\frac{t}{2}cos\frac{t}{3} \, dt$ now apply the second fundamental theorem of calculus (the rule you stated) 3. Originally Posted by th%$&873
I'm struggling with this: $\frac{d}{dx}\int_{x}^{\pi/2} sin\frac{t}{2}cos\frac{t}{3} \, dt$

Using $\frac{d}{dx}\int_{a}^{g(x)} f(t) \, dt =f(g(x)) g'(x)$, I get $\frac{sin\frac{x}{2}sin\frac{x}{3}}{3}-\frac{cos\frac{x}{3}cos\frac{x}{2}}{2}$, which is wrong (I think).
Get ready to kick yourself, sport.

$\frac{d}{dx}\int_{x}^{\pi/2} sin\frac{t}{2}cos\frac{t}{3} \, dt = - \frac{d}{dx}\int_{\pi/2}^{x} sin\frac{t}{2}cos\frac{t}{3} \, dt$.

And you know that $\frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \,$, right?

doh! Beaten by Jhevon. But my answer's funnier

4. Originally Posted by mr fantastic
...
doh! Beaten by Jhevon.
we have a "doh!" emoticon you know:

(the emoticons were a lot better when they were animated)

working on getting nominated for the "Funniest Member" award for 2008 i see

5. Originally Posted by Jhevon
we have a "doh!" emoticon you know: Mr F says: yes, I know. But the emoticons aren't easily accessible (as far as I can see) when you edit a post.

(the emoticons were a lot better when they were animated)

working on getting nominated for the "Funniest Member" award for 2008 i see
To paraphrase the old sporting cliche, I'm just taking it one post at a time

(And if there was a "Grouchiest Member" award, somtimes I think I could easily lose my New Year resolution resolve )

6. apply a minus sign to change the limits
I thought that's what I did--which is why I got $\frac{sin\frac{x}{2}sin\frac{x}{3}}{3}-\frac{cos\frac{x}{3}cos\frac{x}{2}}{2}$ instead of $-\frac{sin\frac{x}{2}sin\frac{x}{3}}{3}+\frac{cos\f rac{x}{3}cos\frac{x}{2}}{2}$
And you know that $\frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \,$, right?
Yes. But why does my calculator give me $-\frac{sin\frac{5x}{6}}{2}-\frac{sin\frac{x}{6}}{2}$?

7. Originally Posted by th%$&873 I thought that's what I did--which is why I got $\frac{sin\frac{x}{2}sin\frac{x}{3}}{3}-\frac{cos\frac{x}{3}cos\frac{x}{2}}{2}$ instead of $-\frac{sin\frac{x}{2}sin\frac{x}{3}}{3}+\frac{cos\f rac{x}{3}cos\frac{x}{2}}{2}$ Mr F says: No no no! You do NOT differentiate f(t) $\frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \,$ NOT $f'(x)$!! Yes. But why does my calculator give me $-\frac{sin\frac{5x}{6}}{2}-\frac{sin\frac{x}{6}}{2}$? Mr F says: Without seeing what your input is, I really couldn't say. But there's an old computer programming adage that has never been truer when it comes to CAS calculators: Garbage in, garbage out. No offence intended .. 8. Originally Posted by mr fantastic ... (And if there was a "Grouchiest Member" award, somtimes I think I could easily lose my New Year resolution resolve ) sorry to burst your bubble, but there are members here who are a lot more grouchy than you are. you may get nominated, but you wouldn't get a single vote...well, maybe if rcmango voted 9. Originally Posted by Jhevon sorry to burst your bubble, but there are members here who are a lot more grouchy than you are. you may get nominated, but you wouldn't get a single vote...well, maybe if rcmango voted Aha yes, but you know I'm holding back My resolve for my New Year resolution stands resolute (currently anyway ) 10. Originally Posted by mr fantastic Mr F says: No no no! You do NOT differentiate f(t) $\frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \, NOT f'(x)!!$ But Jhevon said now apply the second fundamental theorem of calculus (the rule you stated) I would have simply substituted x like you said, mr fantastic, but my calculator said otherwise, so I changed course. Without seeing what your input is, I really couldn't say. But there's an old computer programming adage that has never been truer when it comes to CAS calculators: Garbage in, garbage out. No offence intended Believe me, I am acutely aware of what happens when the input into a calculator isn't formatted right. I entered it exactly as shown in my posting, checked and re-checked. Of course I'm not expecting you to troubleshoot my calculator. 11. Originally Posted by th%$&873
But Jhevon said
I was referring to:
Originally Posted by th%$&873 Using $\frac{d}{dx}\int_{a}^{g(x)} f(t) \, dt =f(g(x)) g'(x)$ that is the right rule to use. it is a variation of the second fundamental theorem of calculus (which mr fantastic gave you the formula for) which is derived by using the chain rule. i was not in anyway referring to the format of your answer. in fact, i was going to ask you in my first post why is it you have two terms as opposed to one 12. got it. 13. Originally Posted by th%$&873
But Jhevon said
I would have simply substituted x like you said, mr fantastic, but my calculator said otherwise, so I changed course.

Believe me, I am acutely aware of what happens when the input into a calculator isn't formatted right. I entered it exactly as shown in my posting, checked and re-checked. Of course I'm not expecting you to troubleshoot my calculator.
If you have a TI-89, the input is:

d( $\int$(sin(t/2)*cos(t/3),t,x, $\pi$/2),x)

and you get ....... ohoho! ..... and you get $-\frac{\sin\frac{5x}{6}}{2}-\frac{\sin\frac{x}{6}}{2}$ ..... now that's interesting .....

But remember that $\sin(A + B) + \sin(A - B) = 2 \sin A \cos B$.

So here's a question for you: What happens if you let $A = -\frac{x}{2}$ and $B = \frac{x}{3}$ in the above identity?

Hint: Stamp this case as solved.

(Or perhaps not ...... why $A = -\frac{x}{2}$ rather than $A = \frac{x}{2}$...... ? Why?)

What's the moral of the story? (The answer has the words "unexpected" and "output" in it). Hmmmm .... the CAS calculator works in mysterious ways sometimes.

14. Originally Posted by th%\$&873
$\frac{d}{dx}\int_{a}^{g(x)} f(t) \, dt =f(g(x)) g'(x)$
And just to seal the deal and end this thing, what does this rule boil down to when g(x) = x?

15. But remember that $\sin(A + B) + \sin(A - B) = 2 \sin A \cos B$.

So here's a question for you: What happens if you let $A = -\frac{x}{2}$ and $B = \frac{x}{3}$ in the above identity?

Hint: Stamp this case as solved.

(Or perhaps not ...... why $A = -\frac{x}{2}$ rather than $A = \frac{x}{2}$...... ? Why?)
My trig is a little rusty.

What's the moral of the story? (The answer has the words "unexpected" and "output" in it). Hmmmm .... the CAS calculator works in mysterious ways sometimes.
That's why I asked you what was going on, because sometimes calculators give odd, but correct results. If you're good at math, you can see what happened. Sometimes the boolean evaluator function doesn't work, and I have to compare two expressions numerically (maybe there's a better way). Bottom line is, I thought this problem was more complicated than it actually was, and got confused.

Originally Posted by mr fantastic
And just to seal the deal and end this thing, what does this rule boil down to when g(x) = x?
$g'(x)
$
is 1, so it would be f(x)?

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