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

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

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

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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. []

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.

member 587159
Mark44
Mentor
@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, 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.