- #1

fishturtle1

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- 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.

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.