Show |x^2 sin^8(e^x)| <= (16 pi^3)/3

  • Thread starter fishturtle1
  • Start date
In summary: I thought that might not have been covered yet.The second thing, you are right. What I showed was just the upper bound. I need to show the lower bound to get the same result. My bad.Also, thanks for the tip on the dx. I will try to be more careful in the future.I guessed that at this point the fundamental theorem of calculus was not yet...well, I thought that might not have been covered yet.The second thing, you are right. What I showed was just the upper bound. I need to show the lower bound to get the same result. My bad.Also, thanks for the tip on the dx. I will try to be more careful in the future.
  • #1
fishturtle1
394
82
Homework Statement
Show ##\vert \int_{-2\pi}^{2\pi} x^2 \sin^8(e^x) \vert \le \frac{16\pi^3}{3}##
Relevant Equations
From previous example, ##\int_{0}^{b} x^2 dx = \frac{b^3}{3}## where ##b > 0##.

##M(f, S) = \sup\lbrace f(x) : x \in S \rbrace##
##m(f, S) = \inf\lbrace f(x) : s \in S \rbrace##

##U(f) = \inf\lbrace M(f, P) : \text{P is a partition of [a, b]} \rbrace##
##L(f) = \sup\lbrace m(f, P) : \text{P is a partition of [a, b]} \rbrace##

We say ##f## is integrable on ##[a,b]## if ##U(f) = L(f)## in which case ##\int_a^b f = U(f)##.

Theorem 33.5: If ##f## is integrable on ##[a,b]##, then ##\vert f \vert## is integrable on ##[a,b]## and $$\vert \int_a^b f \vert \le \int_a^b \vert f \vert$$

Theorem 33.6. Let ##f## be a function defined on ##[a,b]##. If ##a < c < b## and ##f## is integrable on ##[a,c]## and ##[c,b]##, then ##f## is integrable on ##[a,b]## and

$$\int_a^b f = \int_a^c f + \int_c^b f$$
##\textbf{Attempt at solution:}## By theorem 33.5., $$\vert \int_{-2\pi}^{2\pi} x^2\sin^8(e^x) dx \vert \le \int_{-2\pi}^{2\pi} \vert x^2\sin^8(e^x) \vert dx $$ $$= \int_{-2\pi}^{2\pi} \vert x^2 \vert \vert \sin^8(e^x) \vert dx$$ $$\le \int_{-2\pi}^{2\pi} \vert x^2 \vert dx = \int_{-2\pi}^{2\pi} x^2 dx$$

By theorem 33.6, we have

$$\int_{-2\pi}^{2\pi} x^2 dx = \int_{-2\pi}^0 x^2 dx + \int_0^{2\pi}x^2 dx$$

By example 2, we have $$\int_0^{2\pi} x^2 dx = \frac{8\pi^3}{3}$$

Now I try to mimic example 2 to evaluate the other integral, but I'm stuck: Let ##P = \lbrace -2\pi = t_0 < t_1 < \dots < t_n = 0 \rbrace## be a partition of ##[-2\pi, 0]##. We have $$U(x^2, P) = \sum_{k=1}^{n} M(x^2, [t_{k-1}, t_k])\cdot(t_k - t_{k-1}) = \sum_{k=1}^n (t_{k-1})^2\cdot(t_k - t_{k-1})$$

Setting ##t_k = \frac{k(-2\pi)}{n}## we get $$U(f, P) = \sum_{k=1}^n \frac{(k-1)^24\pi^2}{n^2} \frac{(-2\pi)}{n} = -\frac{8\pi^3}{n^3}\sum_{k=1}^n (k-1)^2$$
$$ = -\frac{8\pi^3}{n^3} \cdot \frac 16 (n-1)(n)(2(n-1) + 1) = -\frac{8\pi^3}{n^3}\cdot(2n^3 - 3n^2 + n) \le -\frac{8\pi^3}{3}$$

I was expecting to get ##U(x^2, P) \ge \frac{8\pi^3}{3}## and then could somehow argue ##U(x^2) = \frac{8\pi^3}{3}## but I'm not sure where to go from here.
 
  • Like
Likes member 587159
Physics news on Phys.org
  • #2
Use that $$\int_{-a}^a f = 2\int_0^a f$$
if ##f## is an even function (exercise if you haven't proven already).
 
  • Informative
Likes fishturtle1
  • #3
Math_QED said:
Use that $$\int_{-a}^a f = 2\int_0^a f$$
if ##f## is an even function (exercise if you haven't proven already).

Let ##f## be an even function that is integrable on ##[0,a]##. By definition of even, for all ##x##, ##f(-x) = f(x)##. Let ##S \subset \mathbb{R}## and ##-S = \lbrace -s : s \in S \rbrace##. Then ##\sup\lbrace f(x) : x \in S \rbrace = \sup \lbrace f(x) : x \in -S \rbrace##.

Let ##P = \lbrace 0 = t_0 < t_1 < \dots < t_n = a\rbrace## be any partition of ##[0,a]##. Consider the partition ##-P = \lbrace 0 = -t_0 > -t_1 > \dots > -t_n = -a \rbrace##. Then,

$$U(f, P) = \sum_{k=0}^{n-1} M(f, [t_k, t_k+1]) \cdot (t_{k+1} - t_k)$$
$$ = \sum_{k=0}^{n-1} M(f, [-t_{k+1}, -t_k]) \cdot (-t_k - (-t_{k+1})) $$
$$ = U(x^2, -P)$$

Similarly, for any partition ##Q## of ##[-a, 0]##, we have ##U(f, Q) = U(f, -Q)##. It follows ##\inf\lbrace U(f, P) : \text{P is a partition of [0, a]}\rbrace = \inf\lbrace U(f, Q) : \text{Q is a partition of [-a, 0]} \rbrace##. So ##U(f)## on ##[0,a]## is the same as ##U(f)## on ##[-a, 0]##. By similar reasoning, ##L(f)## on ##[0,a]## is the same as ##L(f)## on ##[-a, 0]##. We can conclude
$$\int_{-a}^a f = \int_{-a}^0 f + \int_0^a f = 2\int_0^a f$$ []

------------

Back to the original problem, (using what we already had in OP), we have

$$\vert \int_{-2\pi}^{2\pi} x^2 \sin^8(e^x) dx \vert \le \int_{-2\pi}^{2\pi} x^2dx = 2 \int_0^{2\pi}x^2dx = 2\cdot \frac{8\pi^3}{3} = \frac{16\pi^3}{3}$$
which is what we wanted. []
 
  • Like
Likes member 587159
  • #4
fishturtle1 said:
Let ##f## be an even function that is integrable on ##[0,a]##. By definition of even, for all ##x##, ##f(-x) = f(x)##. Let ##S \subset \mathbb{R}## and ##-S = \lbrace -s : s \in S \rbrace##. Then ##\sup\lbrace f(x) : x \in S \rbrace = \sup \lbrace f(x) : x \in -S \rbrace##.

Let ##P = \lbrace 0 = t_0 < t_1 < \dots < t_n = a\rbrace## be any partition of ##[0,a]##. Consider the partition ##-P = \lbrace 0 = -t_0 > -t_1 > \dots > -t_n = -a \rbrace##. Then,

$$U(f, P) = \sum_{k=0}^{n-1} M(f, [t_k, t_k+1]) \cdot (t_{k+1} - t_k)$$
$$ = \sum_{k=0}^{n-1} M(f, [-t_{k+1}, -t_k]) \cdot (-t_k - (-t_{k+1})) $$
$$ = U(x^2, -P)$$

Similarly, for any partition ##Q## of ##[-a, 0]##, we have ##U(f, Q) = U(f, -Q)##. It follows ##\inf\lbrace U(f, P) : \text{P is a partition of [0, a]}\rbrace = \inf\lbrace U(f, Q) : \text{Q is a partition of [-a, 0]} \rbrace##. So ##U(f)## on ##[0,a]## is the same as ##U(f)## on ##[-a, 0]##. By similar reasoning, ##L(f)## on ##[0,a]## is the same as ##L(f)## on ##[-a, 0]##. We can conclude
$$\int_{-a}^a f = \int_{-a}^0 f + \int_0^a f = 2\int_0^a f$$ []

------------

Back to the original problem, (using what we already had in OP), we have

$$\vert \int_{-2\pi}^{2\pi} x^2 \sin^8(e^x) dx \vert \le \int_{-2\pi}^{2\pi} x^2dx = 2 \int_0^{2\pi}x^2dx = 2\cdot \frac{8\pi^3}{3} = \frac{16\pi^3}{3}$$
which is what we wanted. []

Well done. I didn't check your entire proof in detail but the idea is correct.
 
  • Wow
  • Like
Likes ush_b9905 and fishturtle1
  • #5
Math_QED said:
Well done. I didn't check your entire proof in detail but the idea is correct.
I appreciate your help.
 
  • Like
Likes member 587159
  • #6
@fishturtle1, you showed way more work than is necessary. Since ##x^2\sin^8(e^x) \ge 0## for all real x, you can immediately discard the absolute value symbols. Here's a much shorter version, using the fact that ##\sin^8(e^x) \ge 0##:
##| \int_{-2\pi}^{2\pi} x^2 \sin^8(e^x)dx| = \int_{-2\pi}^{2\pi} x^2 \sin^8(e^x)dx \le \int_{-2\pi}^{2\pi} x^2 dx##
## = \left.\frac{x^3} 3 \right|_{-2\pi}^{2\pi} = \frac{8\pi^3} 3 - \frac{-8\pi^3} 3 = \frac{16\pi}3##
One more thing -- be sure to include dx in your integral.
 
  • #7
Mark44 said:
@fishturtle1, you showed way more work than is necessary. Since ##x^2\sin^8(e^x) \ge 0## for all real x, you can immediately discard the absolute value symbols. Here's a much shorter version, using the fact that ##\sin^8(e^x) \ge 0##:
##| \int_{-2\pi}^{2\pi} x^2 \sin^8(e^x)dx| = \int_{-2\pi}^{2\pi} x^2 \sin^8(e^x)dx \le \int_{-2\pi}^{2\pi} x^2 dx##
## = \left.\frac{x^3} 3 \right|_{-2\pi}^{2\pi} = \frac{8\pi^3} 3 - \frac{-8\pi^3} 3 = \frac{16\pi}3##
One more thing -- be sure to include dx in your integral.

I guessed that at this point the fundamental theorem of calculus was not yet introduced.
 
  • #8
Mark44 said:
@fishturtle1, you showed way more work than is necessary. Since ##x^2\sin^8(e^x) \ge 0## for all real x, you can immediately discard the absolute value symbols. Here's a much shorter version, using the fact that ##\sin^8(e^x) \ge 0##:
##| \int_{-2\pi}^{2\pi} x^2 \sin^8(e^x)dx| = \int_{-2\pi}^{2\pi} x^2 \sin^8(e^x)dx \le \int_{-2\pi}^{2\pi} x^2 dx##
## = \left.\frac{x^3} 3 \right|_{-2\pi}^{2\pi} = \frac{8\pi^3} 3 - \frac{-8\pi^3} 3 = \frac{16\pi}3##
One more thing -- be sure to include dx in your integral.
Thank you for the response. I agree with what you say, but yes, the fundamental theorem of calculus is in next chapter. The way I have ##\int_0^{2\pi} x^2 dx = \frac{8\pi^3}{3}## is by an example using definition of Riemann integral and specifying a partition.
 

Related to Show |x^2 sin^8(e^x)| <= (16 pi^3)/3

1. What is the purpose of this equation?

The equation "Show |x^2 sin^8(e^x)| <= (16 pi^3)/3" is used to represent a mathematical inequality and is commonly used in calculus and other areas of mathematics to solve various problems.

2. How do I read this equation?

The vertical bars "|" indicate absolute value, "x^2" means x squared, "sin^8" means sine raised to the eighth power, "e^x" means e raised to the power of x, and "pi" represents the mathematical constant pi. The inequality symbol "≤" means "less than or equal to".

3. What is the significance of (16 pi^3)/3 in this equation?

The value (16 pi^3)/3 is the upper bound or maximum value of the expression on the left side of the inequality. This means that the expression on the left side can never be greater than (16 pi^3)/3.

4. How can I solve this inequality?

To solve this inequality, we need to find the values of x that satisfy the inequality. We can do this by using algebraic manipulation, graphing the equation, or using calculus techniques such as finding critical points and using the first or second derivative test.

5. What are the practical applications of this equation?

This equation can be used to solve real-world problems in various fields such as physics, engineering, economics, and more. For example, it can be used to find the maximum value of a function or to determine the range of possible values for a given variable.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
630
  • Calculus and Beyond Homework Help
Replies
1
Views
454
  • Calculus and Beyond Homework Help
Replies
3
Views
439
  • Calculus and Beyond Homework Help
Replies
1
Views
335
  • Calculus and Beyond Homework Help
Replies
3
Views
546
  • Calculus and Beyond Homework Help
Replies
3
Views
491
  • Calculus and Beyond Homework Help
Replies
8
Views
950
  • Calculus and Beyond Homework Help
Replies
1
Views
327
  • Calculus and Beyond Homework Help
Replies
7
Views
924
  • Calculus and Beyond Homework Help
Replies
9
Views
894
Back
Top