Check that $\int_0^1B_tdt$ is $\int_0^1(1-t)dB_t$ and also $B_1-\int_0^1tdB_t$ I am beginning at learning stochastic calculus, and am stuck in a small problem. The chapter I am reading is about Wiener integral, and it gives an example of:
$$
\int^1_0{B(t,\omega)dt}
$$
Since it has already introduced the idea of Wiener integral, and we are able to calculate $\int^1_0{tdB(t,\omega)}$, the author uses integration by parts to obtain the following result:
$$
\int^1_0{B(t,\omega)dt}=B(t,\omega)(t-1)|^1_0-\int^1_0{(t-1)dB(t,\omega)}
=\int^1_0{(1-t)dB(t,\omega)}
$$ 
which is a Gaussian random variable with mean $0$ and variance $1/3$. (The Wiener integral in the last term is the same as that of Riemann Stieltjes integral.)
What I am confused is what happens if $t-1$ is replaced by $t$.
My calculation:
$$
\int^1_0{B(t,\omega)dt}=B(t,\omega)t|^1_0-\int^1_0{tdB(t,\omega)}
=B(1,\omega)-\int^1_0{tdB(t,\omega)}
$$ 
And it turns out to be a Gaussian with variance $1$ minus another Gaussian with variance $1/3$. I am wondering it is the same as the above calculation, or there are errors in my reasoning.

As Did mentioned,  we can use the definition of Ito's Integral.
The process $X(t)=1-t$ is continues deterministic function, so  $X_t$ is cadlag and $\int_{0}^{1}(1-t)dB_t\,$is well define. If $I=\{t_0,t_1,\cdots,t_n\}$ is a sequence of partitions of $[0,1]$  with mesh going to zero, then the Itô integral of $X_t=1-t$ with respect to $B_t$ up to time $t=1$ is a random variable
$$I=\int_{0}^{1}(1-t)dB_t=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{(1-t_{i-1})(B({{t}_{i}})-B({{t}_{i-1}})})\\
\qquad I=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{(B({{t}_{i}})-B({{t}_{i-1}})})-\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{t_{i-1}(B({{t}_{i}})-B({{t}_{i-1}})})
$$
We have
$$I=\underset{n\to \infty }{\mathop{\lim }}(B(t_n)-B(t_0))-\int_{0}^{1}tdB(t)$$
$$I=B(1)-B(0)-\int_{0}^{1}tdB(t)=B(1)-\int_{0}^{1}tdB(t)$$
