integration by parts for numerical definite integrals I am new to calculus, and I am trying to learn integration by parts.
So I have got a question here, and I have no idea what I am doing wrong (i have been looking at it for about 4 hours now..)
The method as I understand it, is as follows: $\int f(x) g^{\prime}(x) d x=f(x) g(x)-\int f^{\prime}(x) g(x) d x$
Now i have a definite integral, evaluated from 0 to 2, and which looks as follows:
$\int 0.5 x \cdot 0.375 x^{2} d x$
I know the answer to be: $\left.0.046875 x^{4}\right|_{0} ^{2}=0.75$, but nomatter how i twist and turn this thing, my answer is wrong.
My last attemt looks like this:
$\begin{array}{l}
f(x)=0.375 x^{2} \\
f^{\prime}(x)=0.75 x \\
g(x)=0.25 x^{2} \\
g^{\prime}(x)=0.5 x
\end{array}$
$0.375 x^{2} \cdot 0.25 x^{2}-\int 0.75 x \cdot 0.25 x^{2}$
$0.375 x^{2} \cdot 0.25 x^{2}- \frac{0.75 x^{2}}{2} \cdot \frac{0.25 x^{3}}{3}$
Which, evaluated at 0 and 2, comes out to be 0.5.
Where am i going wrong?
 A: I'm assuming you were told to use integration by parts here, because you showed that you know the answer via just multiplying the integrand. The error occurs when you take the integral $\int 0.75x \cdot 0.25x^2$.
You can't integrate them separately as you do.
Usually when you're doing integration by parts, your goal is to pick $f$ and $g'$ so that when you apply integration by parts $\int f' g$ becomes simpler. So in your case, you would probably want to chose $f = .5x$ so that when you apply the integration by parts formula $\int f' g $ is very simple because only $g$ still contains a term with $x$ in it. Try this out and see if it makes the problem easier!
This is suggestive of how you apply integration by parts in more complicated examples where it is required. For example $\int x e^x dx$. Here you wouldn't want to choose $f = e^x$ because this doesn't get any simpler when you differentiate.
A: You evaluated the integral
$$\int 0.75 x \cdot 0.25 x^{2}=\frac{0.75 x^{2}}{2} \cdot \frac{0.25 x^{3}}{3}.$$
This is wrong. The right solution is
$$\frac{3x^4}{64}\: (+C).$$
Check:
Here $$0.375 x^{2} \cdot 0.25 x^{2},$$
if $x=2$ is equal to $0.75$ and equal to $0$ if $x=0$. So the final solution is
$$0.75-0=0.75.$$
