{\bf Problem 1.}

$$(cX - ac)(cY - bc)(cZ - bc) = c^3(X-a)(Y-b)(Z-b) = c^3(XYZ - XYb - XZb
+ Xb^2 - aYZ + aYb + abZ - ab^2)$$
This is $c^3XYZ - c^3XYb - c^3XZb + c^3Xb^2 - c^3aYZ + c^3aYb + c^3abZ
- c^3ab^2,$ based on repeated application of the distributive property.

{\bf Problem 2.}

{\bf (a)}
$$\sum_{i=1}^n Ca_i + Cb_i.$$

{\bf (b)}

$$\sum_{i=1}^n\sum_{j=1}^n Ca_ia_j$$

{\bf (c)}

$$\sum_{i=1}^n (c_1a_i + c_2b_i)$$

{\bf Problem 3.}

$$\log_3(81) + 2\log_4(1024^{-1}) = 4 - 2\cdot 5 = -6$$

{\bf Problem 4.}

$$\int_{-1}^1 xdx = \left[ x^2/2 \right]_{-1}^1 = 0.$$

$$\int_{-1}^1 x^2dx = \left[ x^3/3 \right]_{-1}^1 = 1/3 - (-1/3) = 2/3$$

$$\int_0^1 (x-a)^2 dx = \left[ (x-a)^3/3 \right]_0^1=((1-a)^3 + a^3)/3$$

{\bf Problem 5.}

$$\int_0^\infty e^{-2x} dx = \lim_{x\to\infty}(e^{-2x} - e^0) = -1.$$

$$\int_{-\infty}^\infty e^{-2x} dx = \lim_{x\to\infty}(e^{-2x} - e^2x) =
-\infty.$$

{\bf Problem 6.}

$A,$ $B,$ and $C,$ as bounded intervals, are sets. By definition, if
$x\in A \lor B,$ $x\in A\cap B,$ so if $A\supset C$ and $B\supset
C,$ $\forall x\in C\to x\in A\cap B.$

{\bf Problem 7.}

{\bf (a) }
$$\int_{-\infty}^\infty f(x, y) dx = \int_0^y 24x^3y dx = 6y^5,$$
on $y\in[0,1],$ $0$ elsewhere.

$$\int_{-\infty}^\infty f(x, y) dx = \int_x^1 24x^3y dy = \left[
12x^3y^2 \right]_x^1 = 12x^3(1-x^2),$$ on $x\in[0,1],$ $0$ elsewhere.

{\bf (b) }

Integrating over the plane is equivalent to a double integral across
both variables from $-\infty$ to $\infty,$ so we can find it by

$$\int_0^1 6y^5 dy = 1.$$

{\bf Problem 8.}

No.
Let $f(x,y)$ be $2$ on $0<x,y<1,$ $0$ otherwise.

$$f_1(x) = \int_{\bf R} f(x,y) dy = \int_0^1 2 dy = 2.$$
$$f_2(y) = \int_{\bf R} f(x,y) dx = \int_0^1 2 dx = 2.$$
$$\int_{\bf R} f_1(x)dx \int_{\bf R} f_2(y)dy = \int_0^1 2dx\int_0^1 2dy
= 4.$$

$$\int\int_{\bf R^2} f(x,y) = \int_0^1 \int_0^1 2 dxdy = 2 \neq 4.$$

{\bf Problem 9.}

$$\int_0^\infty \exp(-x)\exp(tx) dx = \int_0^\infty \exp((t-1)x) dx$$
converges on $t-1 < 0 \to t < 1.$

{\bf Problem 10.}

$$(1-(-1/2))\sum_{k=0}^{+\infty} (-1/2)^k = \sum_{k=0}^{+\infty}
(-1/2)^k - \sum_{k=1}^{+\infty} (-1/2)^k = 1.$$

Therefore, the original sum is 2/3.

{\bf Problem 11.}

$$f(x) = e^{-x^2/2}$$

$$f(0) = 1, f'(x) = -xe^{-x^2/2} \to f'(0) = 0, f''(x) = -e^{-x^2/2} +
x^2e^{-x^2/2} \to f''(0) = -1.$$

$$f(x) \approx 1 - x^2.$$

{\bf Problem 12.}

{\bf (a)}

$X = (X_1, X_2, X_3)^T, X^T = (X_1, X_2, X_3).$
$$XX^T = X_1^2 + X_2^2 + X_3^2 = X^TX.$$

{\bf (b)}

$$b^TAa = \pmatrix{2&-4}\pmatrix{2&0\cr1&-1\cr}\pmatrix{1\cr0} =
\pmatrix{0}.$$
$$a^Ta^T = \pmatrix{1&0}\pmatrix{2&1\cr 1&-1\cr} = \pmatrix{2&1}$$

\bye