MHB Graphing definite integral functions

[Bueno]

Hi, how are you?

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

$$
b)\int_0^\sqrt{x} e^{t^2}$$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]

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]

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 $$G(x)$$, then, we have $$G(\sqrt{x})$$, so:

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

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

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

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]

Some points to ponder:

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

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

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

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

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

-Dan

 

[MarkFL]

First, can you state the derivative form of the FTOC?

 

[Bueno]

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 $$ G(x) := \int_a^x f(t)dt$$ is differentiable and

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

So, in this case:

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

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

Is this right?

 

[MarkFL]

...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 $$ G(x) := \int_a^x f(t)dt$$ is differentiable and

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

So, in this case:

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

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

Is this right?

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

 

[Bueno]

But where?
I mean, what is the function I should calculate the derivative first?

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

 

[MarkFL]

Suppose we have the function:

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

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

$$h(x)=G(f(x))-G(a)$$

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

$$h'(x)=G'(f(x))f'(x)=g(f(x))f'(x)$$

Can you now apply this result to the given function?

 

[Bueno]

Then, the result would be:

$$e^{x} 1/2\sqrt{x}$$

?

 

[MarkFL]

I would write:

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

then:

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

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]

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 $$(0, +\infty)$$

The second derivative is:

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

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

It seems that the second derivative also does not "make sense" when $$x < 0$$, it's positive when $$x > \frac{1}{2}$$ 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]

The domain of the function is actually $[0,\infty)$. Since we have:

$$\int_a^a f(x)\,dx=0$$, 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 $$\left( 0,\frac{1}{2} \right)$$ and up on $$\left(\frac{1}{2},\infty \right)$$.

Thus, the point of inflection is:

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

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

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]

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]

For $$0\le a$$, if we have:

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

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

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

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:

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

 

[Bueno]

I think things are clear now.

In the other case:

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

The first derivative would be:

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

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]

Yes, you differentiated correctly. :D

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

$$|x|\equiv\sqrt{x^2}$$