Chapter 6.4, Problem 57E

(a)

To determine

H(0) with f being a continuous function.

Answer to Problem 57E

With f being a continuous function,

13y $H\left(0\right)=0$

Explanation of Solution

Given information:

  

  $H\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)}dt$

Where,

f is a continuous function with domain [0,12].

We have

  $H\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)}dt$

Substitute $x=0$ :

  $H\left(0\right)={\displaystyle {\int }_0^{0}f\left(t\right)}dt=0$

Thus,

No area under the curve exists on the interval [0, 0], because this area will have no width.

(b)

To determine

Interval on which H is increasing.

Answer to Problem 57E

H(x) is increasing on [0, 6].

Explanation of Solution

Given information:

  

  $H\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)}dt$

Where,

f is a continuous function with domain [0,12].

H is increasing on interval [a, b],

When

  $H\text{'}\left(x\right)>0$

For

(a,b).

Since

  $H\text{'}\left(x\right)=\frac{d}{dx}{\displaystyle {\int }_0^{x}f\left(t\right)}dt=f\left(x\right)$

And

  $f\left(x\right)>0$

For all

  $x\in \left(0,6\right)$ .

Thus,

H(x) is increasing on [0, 6].

(c)

To determine

Interval on which the graph of H concave up.

Answer to Problem 57E

H is concave up on the interval [9, 12].

Explanation of Solution

Given information:

  

  $H\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)}dt$

Where,

f is a continuous function with domain [0,12].

For H to be concave up,

The second derivative needs to be positive.

Since

  $H\text{'}\left(x\right)=\frac{d}{dx}{\displaystyle {\int }_0^{x}f\left(t\right)}dt=f\left(x\right)$

As well as

  $H\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left[f\left(x\right)\right]=f\text{'}\left(x\right)$

Thus,

  $H\text{'}\text{'}\left(x\right)$ is positive when $f\text{'}\left(x\right)$ is positive.

In other words,

H is concave up when slope of the graph $f\left(x\right)$ is positive.

According to graph,

This occurs when x is between 9 and 12.

Therefore,

H is concave up on the interval [9, 12].

(d)

To determine

Whether H(12) is positive or negative.

Answer to Problem 57E

H(12) is positive.

Explanation of Solution

Given information:

  

  $H\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)}dt$

Where,

f is a continuous function with domain [0,12].

The region between curve and x-axis,

  $H\left(12\right)={\displaystyle {\int }_0^{12}f\left(t\right)dt}$

Since the net area of the region above the x-axis between 0 and 6 is larger than the net area of the region below the x-axis between 6 and 12,

The total region between the curve and x-axis will be positive.

Thus,

H(12) is positive.

(e)

To determine

Point at which H achieves maximum value.

Answer to Problem 57E

H achieves maximum value at $x=6$ .

Explanation of Solution

Given information:

  

  $H\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)}dt$

Where,

f is a continuous function with domain [0,12].

H(x) will achieve its maximum at an endpoint or a value of x.

Such that

  $H\text{'}\left(x\right)=0$

Since

  $H\text{'}\left(x\right)=f\left(x\right)$

And

  $f\left(x\right)=0$

When

  $x=6$ and $x=12$

Then

  $x=6$ and its endpoints of $x=0$ and $x=12$ are possible locations of the maximum.

Since

  $H\text{'}\left(x\right)=f\left(x\right)>0$ on [0, 6]

And

  $H\text{'}\left(x\right)=f\left(x\right)<0$ on [6, 12]

Then

  $x=6$ is the maximum.

Therefore,

H achieves its maximum value at $x=6$ .

(f)

To determine

Point at which H achieves minimum value.

Answer to Problem 57E

H achieves minimum value at $x=0$ .

Explanation of Solution

Given information:

  

  $H\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)}dt$

Where,

f is a continuous function with domain [0,12].

H will achieve its minimum value at an endpoint or critical point.

From Part (e),

The only critical point was at $x=6$ , where H achieved its maximum.

Then

H will achieve its minimum at one of its endpoints at $x=0$ and $x=12$ .

From Part (b),

Since

  $H\text{'}\left(x\right)>0$

For all x between $x=0$ and $x=12$ ,

H is always increasing.

That means

  $H\left(0\right)<H\left(12\right)$ .

Therefore,

H achieves minimum value at $x=0$ .