Derivative of a difficult integral

[E'lir Kramer]

Spivak, 14.1.iii: Derivative of F, where

I have the FTC. The answer is also given,

But I don't know how to find it.

First I want to try a function decomposition:

and from the Chain Rule:

But now I'm stuck. For I do not know an expression for p' or q'. And worse, r' doesn't show up in the given solution. Thanks in advance for your time :).

 

[LCKurtz]

Spivak, 14.1.iii: Derivative of F, where

I have the FTC. The answer is also given,

But I don't know how to find it.

First I want to try a function decomposition:

and from the Chain Rule:

But now I'm stuck. For I do not know an expression for p' or q'. And worse, r' doesn't show up in the given solution. Thanks in advance for your time :).

If you let $f(y) = \int_{8}^{y}\frac{1}{1+t^{2}+\sin^{2}t}dt$ so $F(x) = \int_{15}^xf(y)dy$ what would $F'(x)$ be in terms of $f$?

 

[E'lir Kramer]

Thanks for your reply. When you put it that way, we have F'(x) = f(y) = \int_{8}^{y} \frac{1}{1+t^{2}+sin^{2}t}dt

But the given answer has \int_{8}^{x} \frac{1}{1+t^{2}+sin^{2}t}dt. How can we just change the integrand variable like that?

 

[LCKurtz]

If you let $f(y) = \int_{8}^{y}\frac{1}{1+t^{2}+\sin^{2}t}dt$ so $F(x) = \int_{15}^xf(y)dy$ what would $F'(x)$ be in terms of $f$?

Thanks for your reply. When you put it that way, we have F'(x) = f(y)

No it doesn't! That y is a dummy variable and F is a function of x. And its derivative must be a function of x.

 

[Ray Vickson]

Thanks for your reply. When you put it that way, we have F'(x) = f(y) = \int_{8}^{y} \frac{1}{1+t^{2}+sin^{2}t}dt

But the given answer has \int_{8}^{x} \frac{1}{1+t^{2}+sin^{2}t}dt. How can we just change the integrand variable like that?

The y inside the y-integral is "integrated over", so goes away. For example, you can think of $F(x) = \int_8^x f(y) \, dy$ as a limit of the form
\lim_{n \to \infty} \sum_{i=1}^{n} \frac{x-8}{n} f\left(8 + \frac{i (x-8)}{n}\right),
and there is no 'y' at all in this expression. In other words,
F(x) = \int_8^x f(\text{anything}) \; d\: \text{anything},
and it does not matter what name you give to "anything" (except that it should not be x, f or d).

 

[E'lir Kramer]

Thanks. I think I get it. Because we're integrating over $dy$, that means we're plugging in $x$ for y when we integrate.

 

[tt2348]

Just remember... \int _{ a }^{ b }{ f(x)dx } =F(b)-F(a) so let's call \int _{ 8 }^{ y }{ g(t)dt } =G(y)-G(8) where g(t)=\frac { 1 }{ 1+ t^{2}+sin^{2}t } then \int _{ 15 }^{ x }{ \int _{ 8 }^{ y }{ g(t)dt } } =\int _{ 15 }^{ x }{ G(y)-G(8){dy} } since the last term is a constant, we get F(x)=\int _{ 15 }^{ x }{ \int _{ 8 }^{ y }{ g(t)dt } } =\left[ H(x)-H(15) \right] -G(8)\left( x-15 \right) =H(x)-G(8)x+15G(8)-H(15) Now derive and see what constants drop out :-)

 

[E'lir Kramer]

Ok guys, I struggled through that one and I think I understand it now. To test my knowledge, I've attempted another one, which isn't solved.

$F(x) = \sin \left ( \int_{0}^{x} \sin \left ( \int_{0}^{y} \sin^{3} t dt \right ) dy \right )$

So now I will try another composition. I will use capital letters for functions defined by integrals, and lowercase for functions that aren't.

$f(x) = \sin x \\ G(x) = \int_{0}^{x} \sin y dy\\ H(y) = \int_{0}^{y} \sin^{3}t dt \\$
then,

$F(x) = f \circ G \circ H$

By the chain rule,

$F'(x) = f'(G \circ H)\cdot G'(H) \cdot H'$

$f'(x) = \cos x$

By the FTC,

$G'(x) = \sin x \\ H'(y) = \sin^{3} y \\$

Composing,

$F'(x) = \cos(\int_{0}^{x} \sin \left ( \int_{0}^{y} \sin^{3}t dt \right )dy \cdot \sin \left (\int_{0}^{x} \sin^{3} t dt \right ) \cdot sin^{3} x$

How'd I do?

 

[Dick]

Ok guys, I struggled through that one and I think I understand it now. To test my knowledge, I've attempted another one, which isn't solved.

$F(x) = \sin \left ( \int_{0}^{x} \sin \left ( \int_{0}^{y} \sin^{3} t dt \right ) dy \right )$

So now I will try another composition. I will use capital letters for functions defined by integrals, and lowercase for functions that aren't.

$f(x) = \sin x \\ G(x) = \int_{0}^{x} \sin y dy\\ H(y) = \int_{0}^{y} \sin^{3}t dt \\$
then,

$F(x) = f \circ G \circ H$

By the chain rule,

$F'(x) = f'(G \circ H)\cdot G'(H) \cdot H'$

$f'(x) = \cos x$

By the FTC,

$G'(x) = \sin x \\ H'(y) = \sin^{3} y \\$

Composing,

$F'(x) = \cos(\int_{0}^{x} \sin \left ( \int_{0}^{y} \sin^{3}t dt \right )dy \cdot \sin \left (\int_{0}^{x} \sin^{3} t dt \right ) \cdot sin^{3} x$

How'd I do?

You are missing a big closing parentheses on the first cos( part. But it looks ok to me.

 

[LCKurtz]

I don't think it is quite correct. If you call$$
f(y) = \sin(\int_0^y \sin^3(t)dt)$$ then$$
F(x) = \sin(\int_0^xf(y)dy)$$Then$$
F'(x) = cos(\int_0^xf(y)dy)\cdot \frac d {dx}\int_0^xf(y)dy=
cos(\int_0^xf(y)dy)\cdot f(x)
=cos(\int_0^xf(y)dy)\cdot \sin(\int_0^x \sin^3(t)dt)$$ $$
=cos(\int_0^x\sin(\int_0^y \sin^3(t)dt)dy)\cdot \sin(\int_0^x \sin^3(t)dt)
$$I think you have an extra $\sin^3(x)$.

 

[Dick]

Yeah, you are right. Ooops.

 

[E'lir Kramer]

Though I am convinced by Kurtz's answer, I am having a hard time seeing where I went wrong in my own reasoning.

Is my function composition wrong? Does $F(x) ≠ (f \circ G \circ H)(x)$?

 

[LCKurtz]

Though I am convinced by Kurtz's answer, I am having a hard time seeing where I went wrong in my own reasoning.

Is my function composition wrong? Does $F(x) ≠ (f \circ G \circ H)(x)$?

You have $G(x) = \int_0^x \sin y\, dy$, but let's call that dummy variable of integration $u$ so $G(x) = \int_0^x \sin u\, du$. Also, you have $H(y) = \int_0^y \sin^3 t\, dt$. Then$$
G(H(y)) =\int_0^{H(y)} \sin u\, du=\int_0^{\int_0^y \sin^3 t\, dt} \sin u\, du$$I don't think that is what you want.