True or False Integral Calculus Question #1

[MermaidWonders]

True or False: If $$h(t) > 0$$ for $$0 \le t\le 1$$, then the function $$H(x) = \int_{0}^{x} h(t)\,dt$$ is concave up for $$0 \le t\le 1$$.

 

[MarkFL]

Suppose $h(t)=c$, where $0<c$...what is $H(x)$?

 

[MermaidWonders]

If $$c$$ is a constant, then $$H(x)$$ = $$cx$$?

 

[tkhunny]

What is the definitive test for "Concave Up"?

Can you apply this test to your function?

 

[MermaidWonders]

Like where a function's second derivative is > 0?

 

[Greg]

Hi MermaidWonders and welcome to MHB!

For inline latex, use \$...\$.

For example, \$x^2=4\$ gives $x^2=4$, inline.

 

[I like Serena]

Like where a function's second derivative is > 0?

Yep.
From wiki:

f is called convex (or convex downward or concave upward) if the line segment between any two points on the graph of the function lies above or on the graph.​

and:

If f is twice-differentiable, then f is convex if and only if f'' is non-negative. If its second derivative is positive then it is strictly convex.​

What is $H''(x)$?

 

[MermaidWonders]

Wait, so how do I get the sign of $H''(x)$? From the question, I know that since the fundamental theorem tells me that $H'(x)$ = $h(x)$, and I'm given that $h(t) > 0$, $H'(x)$ will also be $> 0$. But how do I figure out the concavity from there, knowing that $H'(x) > 0$?

 

[I like Serena]

Wait, so how do I get the sign of $H''(x)$? From the question, I know that since the fundamental theorem tells me that $H'(x)$ = $h(x)$, and that $h(t) > 0$, $H'(x)$ will also be $< 0$. But how do I figure out the concavity from there?

Indeed. So we know that $H'(x)>0$ (positive not negative) for $0\le x \le 1$, but we still don't know anything about $H''(x)$.
We can only tell that $H$ is strictly increasing and that it is strictly positive on the interval. We cannot say if it's convex or concave.

 

[MarkFL]

Indeed. So we know that $H'(x)>0$ (positive not negative) for $0\le x \le 1$, but we still don't know anything about $H''(x)$.
We can only tell that $H$ is strictly increasing and that it is strictly positive on the interval. We cannot say if it's convex or concave.

Indeed, and is why I resorted to looking for a function satisfying the given criteria, that is not concave up on the given interval. :)

 

[I like Serena]

Indeed, and is why I resorted to looking for a function satisfying the given criteria, that is not concave up on the given interval. :)

Unfortunately, with $h(t)=c$, the function $H$ is actually concave up ($H''(x)\ge 0$), just not strictly concave up.
But yes, it does illustrate that we cannot just make the function (strictly) concave up.

 

[MarkFL]

Unfortunately, with $h(t)=c$, the function $H$ is actually concave up ($H''(x)\ge 0$), just not strictly concave up.
But yes, it does illustrate that we cannot just make the function (strictly) concave up.

Perhaps there are differences in terminology...I was taught that a function is concave up on an interval when its second derivative is positive, and concave down when its second derivative is negative. When the derivative of a function is constant, then the function itself has no concavity.

 

[MermaidWonders]

Ah, so if we don't know if it's concave up or concave down, the statement is false?

 

[I like Serena]

Ah, so if we don't know if it's concave up or concave down, the statement is false?

Correct. (Nod)

 

[MermaidWonders]

Wait, I'm still confused. Can anyone give me like an overall explanation/walkthrough to get to this conclusion? I really want to make sure I understand it. :(

 

[I like Serena]

Wait, I'm still confused. Can anyone give me like an overall explanation/walkthrough to get to this conclusion? I really want to make sure I understand it. :(

With the given information we can conclude that $H'(x)>0$ on the unit interval due to the Fundamental Theorem of Calculus, meaning it's strictly increasing.
However, we have no information whatsoever on $H''$, meaning we cannot say anything about whether it's concave up or not.
So the statement could be true in some cases, but it is not true for the general case.
Proof
$h(t)=2-t^2$ is a counter example. We have $h(t)>0$ for $0\le t \le 1$, but $H''(x)=h'(x)=-2x < 0$ for $0<x\le 1$.
It means that the statement is false.

 

[I like Serena]

Perhaps there are differences in terminology...I was taught that a function is concave up on an interval when its second derivative is positive, and concave down when its second derivative is negative. When the derivative of a function is constant, then the function itself has no concavity.

I can only refer to the wiki page, which is consistent with what I've learned. And I've never heard or seen that it could be different. Can you perhaps provide a reference?

 

[MarkFL]

Wait, I'm still confused. Can anyone give me like an overall explanation/walkthrough to get to this conclusion? I really want to make sure I understand it. :(

Let's look at an example that more clearly demonstrates the statement regarding concavity isn't always true. Consider:

$$H(x)=\ln(x+1)$$

And so:

$$H'(x)=h(x)=\frac{1}{x+1}$$

We have:

$$h(t)>0$$ for $0\le t\le1$

But:

$$H''(x)=-(x+1)^{-2}$$

On the given interval, this function is concave down, but it meets the given criteria, so we know the statement is false.

 

[MarkFL]

I can only refer to the wiki page, which is consistent with what I've learned. And I've never heard or seen that it could be different. Can you perhaps provide a reference?

I was simply taught that a linear function has no concavity...just recalling from memory. :)

 

[I like Serena]

I was simply taught that a linear function has no concavity...just recalling from memory. :)

I think we have no concavity if $f''(x)=0$ for all $x$ in the domain.
However, if we have $f''(x)=0$ only on part of the domain, the function can still be convex or concave.
For instance $f(x)=x^4$ counts as a concave upwards function even though $f''(0)=0$.
Moreover, it's even strictly concave upwards, since any line segment is still strictly above the graph, which is the actual definition (also explained in the wiki article).

 

[MermaidWonders]

Oh, I see. I finally get it now. Many thanks to all those who have helped me with this question! :)