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Graphing definite integral functions

Bueno

New member
Mar 29, 2013
14
Hi, how are you?

I came across some exercises that really puzzled me. They ask me to graph the following functions:
\(\displaystyle
a) \int_0^x\sqrt{|tan(w)|} dw
\)

\(\displaystyle
b)\int_0^\sqrt{x} e^{t^2}\)


I imagine I'll have to use derivative techniques as I would when graphing a "normal" function, but those integral signs and intervals are very confusing.

What should I do?

Thank you,

Bueno
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, analysis of the first and second derivatives (along with a root) will allow you to sketch a reasonable graph of the functions. Can you apply the derivative form of the fundamental theorem of calculus to find the first derivatives?
 

Bueno

New member
Mar 29, 2013
14
I thought of something like this:

I don't know if I really have to take into account the limits of integration to find the derivative of this function. If I have to, I thought of something about the upper limit:

This integral is a function \(\displaystyle G(x)\), then, we have \(\displaystyle G(\sqrt{x})\), so:

\(\displaystyle G'(\sqrt{x}) = (1/2\sqrt{x}) G'(\sqrt{x})
\)

Since \(\displaystyle G(x) = e^{t^2}, then\)

\(\displaystyle G'(x) = 2te^{t^2}
\)
So:
\(\displaystyle

G'(\sqrt{x})= (1/2\sqrt{x}) 2\sqrt{x}e^{x} = e^{x} \)

But I'm not sure about the lower (0).

These limits of integration really confuse me.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
Some points to ponder:

This integral is a function \(\displaystyle G(x)\), then, we have \(\displaystyle G(\sqrt{x})\), so:

\(\displaystyle G'(\sqrt{x}) = (1/2\sqrt{x}) G'(\sqrt{x})
\)
The G'(sqrt(x)) cancels on both sides of this equation!

Since \(\displaystyle G(x) = e^{t^2}, then\)
LHS is a function of x, RHS is a function of t...

-Dan
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
First, can you state the derivative form of the FTOC?
 

Bueno

New member
Mar 29, 2013
14
Some points to ponder:


The G'(sqrt(x)) cancels on both sides of this equation!


LHS is a function of x, RHS is a function of t...

-Dan
What a mess!
I think I "mixed" some Xs and Ts during the calculations.

First, can you state the derivative form of the FTOC?
I've never hear about this form of the FTOC, but after some researches I came across this definition:


"Let f be a continuous function and let a be a constant.
Then the function \(\displaystyle G(x) := \int_a^x f(t)dt\) is differentiable and

\(\displaystyle G'(x) = d/dx (\int_a^x f(t)dt) = f(x)\)

So, in this case:

\(\displaystyle G(x) =\int_0^\sqrt{x} e^{t^2}\)

Then: \(\displaystyle G'(x) = e^{\sqrt{x}^2} = e^{x}\)

Is this right?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
...I've never hear about this form of the FTOC, but after some researches I came across this definition:


"Let f be a continuous function and let a be a constant.
Then the function \(\displaystyle G(x) := \int_a^x f(t)dt\) is differentiable and

\(\displaystyle G'(x) = d/dx (\int_a^x f(t)dt) = f(x)\)

So, in this case:

\(\displaystyle G(x) =\int_0^\sqrt{x} e^{t^2}\)

Then: \(\displaystyle G'(x) = e^{\sqrt{x}^2} = e^{x}\)

Is this right?
Not quite...you must apply the chain rule in this case.
 

Bueno

New member
Mar 29, 2013
14
But where?
I mean, what is the function I should calculate the derivative first?

In the upper limit of integration (\(\displaystyle \sqrt{x}\))?
Or \(\displaystyle e^{t^2}\) ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Suppose we have the function:

\(\displaystyle h(x)=\int_a^{f(x)}g(t)\,dt\) where $a$ is a constant.

If \(\displaystyle G'(x)=g(x)\), then we may use the anti-derivative form of the fundamental theorem of calculus to state:

\(\displaystyle h(x)=G(f(x))-G(a)\)

Now, differentiating both sides with respect to $x$, we find:

\(\displaystyle h'(x)=G'(f(x))f'(x)=g(f(x))f'(x)\)

Can you now apply this result to the given function?
 

Bueno

New member
Mar 29, 2013
14
Then, the result would be:

\(\displaystyle e^{x} 1/2\sqrt{x}\)

?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would write:

Given \(\displaystyle f(x)=\int_0^{\sqrt{x}}e^{t^2}\,dt\)

then:

\(\displaystyle f'(x)=e^{(\sqrt{x})^2}\frac{d}{dx}(\sqrt{x})=\frac{e^x}{2\sqrt{x}}\)

Now, can you find the point $(0,f(0))$, describe the function's behavior on its domain, including increasing/decreasing intervals, and concavity, including any inflection points?
 

Bueno

New member
Mar 29, 2013
14
It does not have any real root, so there are no critical points.
It doesn't make sense for x < 0 and is always positive in it's domain (which seems to be \(\displaystyle (0, +\infty)\)

The second derivative is:

\(\displaystyle \frac{e^{x} (2x - 1)}{4x^{3/2}}\)

What gives us the inflection point \(\displaystyle (\frac{1}{2}; 0)\)

It seems that the second derivative also does not "make sense" when \(\displaystyle x < 0\), it's positive when \(\displaystyle x > \frac{1}{2}\) and negative between 0 and 1/2.

But I have a question:

The lower limit of integration doesn't affect directly the calculation of the first derivative because it's a constant, right?
But does it affect the domain of the function I want to plot?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The domain of the function is actually $[0,\infty)$. Since we have:

\(\displaystyle \int_a^a f(x)\,dx=0\), we know the origin is the left end-point of the function.

As you correctly observed, the first derivative has no real roots, and since is is non-negative on the entire domain, we know the function is strictly increasing on the entire domain.

Your second derivative is correct. We know then that concavity is down on \(\displaystyle \left( 0,\frac{1}{2} \right)\) and up on \(\displaystyle \left(\frac{1}{2},\infty \right)\).

Thus, the point of inflection is:

\(\displaystyle \left(\frac{1}{2},f\left(\frac{1}{2} \right) \right)\approx(0.5,0.544987)\)

I used numeric integration to obtain an approximate value for the function at \(\displaystyle x=\frac{1}{2}\).

You are correct that the lower limit of integration does not affect the calculation of the derivative, but it does in essence tell us where the function "begins."
 

Bueno

New member
Mar 29, 2013
14
I'm sorry, but I didn't understand what you mean by "where the function starts".

When I'm plotting this graph, the lower limit tells me where I should start?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
For \(\displaystyle 0\le a\), if we have:

\(\displaystyle f(x)=\int_a^{\sqrt{x}}e^{t^2}\,dt\)

then we know the point $(a^2,0)$ is on the curve, and we know the left end-point is:

\(\displaystyle \left(0,-\frac{\sqrt{\pi}}{2}\text{erfi}(a) \right)\)

So, the lower limit of integration serves to act as a vertical shift on the function.

When $a<0$, then there is no $x$-intercept, and the left end-point is:

\(\displaystyle \left(0,\frac{\sqrt{\pi}}{2}\text{erfi}(-a) \right)\)
 

Bueno

New member
Mar 29, 2013
14
I think things are clear now.

In the other case:

\(\displaystyle \int_0^x\sqrt{|tan(w)|} dw\)

The first derivative would be:

\(\displaystyle \frac{d}{dx} \int_0^x\sqrt{|tan(w)|} dw = \sqrt{|tan(x)|} \)

Am I right?

If so, how to calculate the second derivative?
When I came across other functions involving absolute value, they were polynomial so I could determine where they would change signs or not, but in this case there are infinite intervals where I would have to change signs.
Probably I'm missing something, but I can't realize what it is.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, you differentiated correctly. :D

I sometimes find it useful when differentiating a function with absolute values to use the definition:

\(\displaystyle |x|\equiv\sqrt{x^2}\)