The images you uploaded seem to show two parts of a mathematical problem involving piecewise functions and integrals.

1. First Image:

   • You have the integral of a piecewise function $f(x)$. It’s split into two cases:
     • $f(x) = \left( \frac{3x}{2} + 1 \right)^9$ when $x < 0$
     • $f(x) = x \cdot \sin(x)$ when $x \geq 0$
   • The goal is to compute the integral from $-2$ to $2$.

   We can break this integral into two parts, one for each piece of the function: $\int_{-2}^2 f(x) \, dx = \int_{-2}^0 \left( \frac{3x}{2} + 1 \right)^9 \, dx + \int_0^2 x \cdot \sin(x) \, dx$ We’ll calculate both integrals separately.

2. Second Image:

   • This image talks about using integration by parts to compute the second integral $I_2 = \int_0^2 x \sin(x) \, dx$.
   • The integration by parts formula is: $\int u \, dv = uv - \int v \, du$
   • To apply this to $I_2$, we choose:
     • $u(x) = x$, hence $u'(x) = 1$
     • $v'(x) = \sin(x)$, hence $v(x) = -\cos(x)$

   Applying the integration by parts formula: $\int_0^2 x \sin(x) \, dx = \left[ -x \cos(x) \right]_0^2 + \int_0^2 \cos(x) \, dx$ $= \left( -2 \cos(2) + 0 \right) + \left[ \sin(x) \right]_0^2$ $= -2 \cos(2) + \sin(2)$ This is the value for $I_2$.

Next, I will compute the integral for $I_1 = \int_{-2}^0 \left( \frac{3x}{2} + 1 \right)^9 \, dx$, and combine both results.

Let me know if you'd like to continue with these computations or need further clarification!