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

[fishturtle1]

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.

 

[member 587159]

Use that $$\int_{-a}^a f = 2\int_0^a f$$
if $f$ is an even function (exercise if you haven't proven already).

 

[fishturtle1]

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
$$\int_{-a}^a f = \int_{-a}^0 f + \int_0^a f = 2\int_0^a f$$ []

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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. []

 

[member 587159]

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.

 

[fishturtle1]

Well done. I didn't check your entire proof in detail but the idea is correct.

I appreciate your help.

 

[Mark44]

@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.

 

[member 587159]

@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.

 

[fishturtle1]

@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.