Help starting with Integration by parts I am working on an assignment that deals with integration by parts and u-substitution and I have become frustratingly stuck on this question. I am unsure where to even start.
Question:
Let $f$ be a twice differentiable function such that $f(0)=5$, $f(3)=1$, and $f^{'}(3)=-2$. Determine the value of $$\int_0^3xf^{''}(x)dx$$
I would greatly appreciate if someone could explain to me at least how to start this.
Thanks.
 A: Let $u=x$ and $dv=f''(x)\,dx$. Then $du=dx$ and we can take $v=f'(x)$. So our integral is equal to
$$\left.xf'(x)\right|_0^3-\int_0^3 f'(x)\,dx.$$
Note that the remaining integral is $f(3)-f(0)$.  
A: Let $v=f'(x)$ and $u=x$. Then $dv=f''(x)dx$ and $du=dx$. Integrating by parts yield
$$\int_0^3xf''(x)dx=\left.xf'(x)\right|_0^3-\int_0^3f'(x)dx=3(-2)-0-\left.f(x)\right|_0^3=-6-(1-5)=-2$$
A: Using Integration by Parts, we can do the following:
Let $u=x, \ dv = f^{\prime\prime}(x)dx \implies du = dx ,\ v=f^\prime(x)$.
Now \begin{align*}\int\limits_{0}^{3}xf^{\prime\prime}(x)dx &= [xf^{\prime}(x)]^3_0 - \int\limits_{0}^3 f^{\prime}(x)dx \\ &= 3(-2) - [f(x)]^3_0 \\&=-6 - (1 -5)  \\ &=-6+4\\ &=-2\end{align*}
A: When I was first learning Substitution by Parts, the following approach was very useful to me in trying to understand the method.
By the multiplication rule for derivatives.
$$f'(x)G(x)+f(x)G'(x)=[f(x)G(x)]'$$
$$f(x)G'(x)=[f(x)G(x)]'-f'(x)G(x)$$
Let $g(x) = G'(x)$, then
$$f(x)g(x)=[f(x)\bigg(\int g(x)dx\bigg)]'-f'(x)\bigg(\int g(x)dx\bigg)$$
$$\int f(x)g(x)dx=f(x)\bigg(\int g(x)dx\bigg)-\int f'(x)\bigg(\int g(x)dx\bigg)dx$$
This shows me that when integrating the product of two functions that one factor will be differentiated $(f)$ and the other will be integrated $(g)$. Technically this equation (as written) isn't correct because it assumes that both $\int g(x)dx$'s evaluate to the same function (e.g. the first $\int g(x)dx$ could evaluate to $h(x)$, and the other one to $h(x)+C$), but it provides a nice direct way of thinking about the method (i.e. it allows me to avoid asking convoluted questions like "$g$ is the derivative of what function?"). To make it technically correct, replace both $\int g(x)dx$'s with $G(x)$.
Applying this point of view to your problem we have
$$\begin{array}{lll}
\int^{3}_{0}xf''(x)dx&=&\bigg[x\bigg(\int f''(x)dx\bigg)-\int 1\cdot \bigg(\int f''(x)dx\bigg)dx \bigg]^3_0\\
&=&\bigg[xf'(x)-\int f'(x)dx\bigg]^3_0\\
&=&\bigg[xf'(x)-f(x)\bigg]^3_0\\
&=&3f'(3)-f(3) -[0f'(0)-f(0)]\\
&=&3(-2)-1 +5 = -2\\
\end{array}$$
As an example, try it with
$$\int \ln x dx = \int (\ln x)(1) dx=\dots$$
