# Graphing definite integral functions

#### Bueno

##### New member
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

Staff member
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
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
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

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

#### Bueno

##### New member
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

Staff member
...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
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

Staff member
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
Then, the result would be:

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

?

#### MarkFL

Staff member
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
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

Staff member
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
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

Staff member
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
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

Yes, you differentiated correctly. $$\displaystyle |x|\equiv\sqrt{x^2}$$