Split the integral at $x = 0$ where $|x|$ changes definition:

$$\int_{-1}^2 |x|\, dx = \int_{-1}^0 (-x)\, dx + \int_0^2 x\, dx$$

Evaluating each part:

$$\int_{-1}^0 (-x)\, dx = \left[ -\frac{x^2}{2} \right]_{-1}^0 = 0 - \left(-\frac{1}{2}\right) = \frac{1}{2}$$ $$\int_0^2 x\, dx = \left[ \frac{x^2}{2} \right]_0^2 = \frac{4}{2} - 0 = 2$$

Total value:

$$I = \frac{1}{2} + 2 = \frac{5}{2}$$

Expand $\sin 2x$ using Maclaurin series:

$$\sin 2x = 2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \dots = 2x - \frac{8x^3}{6} + O(x^5) = 2x - \frac{4}{3}x^3 + O(x^5)$$

Substitute into the given expression:

$$\frac{\sin 2x + ax + bx^3}{x^3} = \frac{(2x - \frac{4}{3}x^3 + O(x^5)) + ax + bx^3}{x^3}$$ $$= \frac{(2 + a)x + (b - \frac{4}{3})x^3 + O(x^5)}{x^3} = \frac{2 + a}{x^2} + \left(b - \frac{4}{3}\right) + O(x^2)$$

For the limit as $x \to 0$ to exist and be finite:

• The coefficient of $\frac{1}{x^2}$ must be zero:
  $$2 + a = 0 \implies a = -2$$
• The remaining constant term must equal the limit value 1:
  $$b - \frac{4}{3} = 1 \implies b = 1 + \frac{4}{3} = \frac{7}{3}$$

A function is concave up on intervals where its second derivative is strictly positive ($f''(x) \gt 0$):

• First derivative:
  $$f'(x) = 4x^3 - 12x^2$$
• Second derivative:
  $$f''(x) = 12x^2 - 24x = 12x(x - 2)$$
• Set $f''(x) \gt 0$:
  $$12x(x - 2) \gt 0$$
  The roots are $x = 0$ and $x = 2$. The product is positive when:
  $x \lt 0$ or $x \gt 2$
  In interval notation, the graph is concave up on $(-\infty, 0) \cup (2, \infty)$.