Volume calculation with n-variables integrals Given $A=(a_{i.j})_{1\le i,j \le n}$ invertible matrix of size $n \times n$, and given $T$ the domain in $\mathbb{R}^n$ which is defined by the following inequality: $\alpha_i \le \sum_{j=1}^{n}{a_{i,j}x_j} \le \beta_i$.
(a). How can I calculate the volume $V(T)$?
(b). Given $f(x)=\sum_{i=1}^n{c_i x_i}$ for constants $c_1,...,c_n$, How can I proove that $\int_T{fd_T}=\frac{V(T)}{2}\sum_{i=1}^{n}{d_i(\beta_i + \alpha_i)}$.   $d_1,...,d_n$ are constants.
For (a): I tried to substitute the variables- $x_i$ becomes $y_i$ so that $y=Bx$ ($B$ is the derivatives matrix by each variable), and then I stucked with the Jacobian calculation, but I think that the integral will be $V(T)=(\beta_1 - \alpha_1)(\beta_2 - \alpha_2)...(\beta_n - \alpha_n)$
Thank you!

Note that $x \in T$ iff $Ax \in \prod_{i=1}^n [\alpha_i, \beta_i]$, that is $AT = \prod_{i=1}^n [\alpha_i, \beta_i]$ so we have, by the integral transformation formula 
\begin{align*}
  V(T) &= \int_T \,dx\\
       &= \int_{AT} |\det A^{-1}|\, dy \\
       &= \frac 1{|\det A|} \cdot \prod_{i=1}^n (\beta_i - \alpha_i)
\end{align*}
To integrate the linear $f$ we argue along the same lines
\begin{align*}
  \int_T f(x)\, dx &=  \int_{AT} f(A^{-1}y)\cdot |\det A^{-1}|\, dy\\
\end{align*}
Now note, that $y \mapsto f(A^{-1}y)$ is linear as a composition of two linear functions, so there are $d_i$ such that $f(A^{-1}y) = \sum_i d_iy_i$ for each $y$. This gives
\begin{align*}
  \int_T f(x)\, dx &=  \int_{AT} f(A^{-1}y)\cdot |\det A^{-1}|\, dy\\
      &= \frac 1{|\det A|} \int_{\prod_j [\alpha_j, \beta_j]} \sum_i d_iy_i\, dy\\
     &= \sum_i  \frac {d_i}{|\det A|} \int_{\prod_j [\alpha_j, \beta_j]} y_i\, dy\\
     &= \sum_i \frac{d_i}{|\det A|} (\beta_1 - \alpha_1) \cdots (\beta_{i-1}-\alpha_{i-1}) \cdot (\beta_{i+1} - \alpha_{i+1}) \cdots (\beta_n -\alpha_n) \cdot \int_{\alpha_i}^{\beta_i} y_i\, dy_i\\
    &= \sum_i \frac{d_i}{|\det A|} (\beta_1 - \alpha_1) \cdots (\beta_{i-1}-\alpha_{i-1}) \cdot (\beta_{i+1} - \alpha_{i+1}) \cdots (\beta_n -\alpha_n) \cdot \frac{\beta_i^2 - \alpha_i^2}2\\
    &= \sum_i \frac{d_i}{2|\det A|} \prod_j (\beta_j - \alpha_j) \cdot (\beta_i + \alpha_i)\\
    &= \frac{V(T)}2 \cdot \sum_i d_i(\alpha_i + \beta_i)
\end{align*}
