Question

Help please? If f is the antiderivative of (x^2/(1+x^5)) such that f(1) = 0 then f(4) =?

Answer 1

I understand this is integration but i do not know how to do it sadly:(

Answer 2

\[\int\limits_{1}^{4} \frac{ x^2 }{ 1+x^5 } dx\] I guess this is what you have to integrate

Answer 3

ooh so the 1 and 4 are the starting and stopping values... but how do you integrate it?

Answer 4

Haha, well it's a bit complicated

Answer 5

Is there a general formula i can follow?

Answer 6

You can try checking to see if \(x^5+1\) has any real roots, and maybe get some partial fractions going...

Answer 7

Siths is much better at this stuff than me haha, I'll let him take over :P

Answer 8

Ok! Sith would the real roots be like x = -1 and x = -1^2/5?

Answer 9

\[x^5+1=(x+1)(x^4-x^3+x^2-x+1)\]
so then
\[\begin{align*}\frac{x^2}{x^5+1}&=\frac{A}{x+1}+\frac{Bx^3+Cx^2+Dx+E}{x^4-x^3+x^2-x+1}\\\\
x^2&=(A+B)x^4+(-AB+C)x^3+(A+C+D)x^2\\
&\quad+(-A+D+E)x+(A+E)
\end{align*}\]

Answer 10

There's only one real root to consider here, the other four are complex.

Answer 11

I think we can approximate it with simpsons rule as well?

Answer 12

@iambatman we're not computing a definite integral here, but rather solving a differential equation. The given info, that \(f(1)=0\), is an initial condition.

Once we compute the antiderivative, we have an unknown constant, i.e. when we get a solution like
\[f(x)=(\text{some expression in terms of }x)+C\]
we plug in the initial condition and get
\[0=(\text{expression with }x\text{ replaced with }1\text{s})+C\]
and we can solve for \(C\).

Answer 13

Ah, gotcha :)

Answer 14

ok so i understand the whole general expression + c but i do not know why you did the previous breakdown of the equation with the ABCs

Answer 15

It's the partial fraction decomposition. We have some rational function of the form \(\dfrac{P(x)}{Q(x)}\), where \(P\) and \(Q\) are polynomials. We break up \(Q(x)\) into irreducible factors \(S_1(x),\ldots,S_n(x)\) (also polynomials) such that
\[\frac{P(x)}{Q(x)}=\sum_{k=1}^n\frac{R_k(x)}{S_k(x)}\]
where \(R_k(x)\) is one degree less than each corresponding \(S_k(x)\).

Basically what this means is that we can split up a rational function into more convenient rational functions.

It's the reverse of the process shown here, for instance:
\[\frac{x}{x^2+1}+\frac{1}{x-1}=\frac{x(x-1)+(x^2+1)}{(x^2+1)(x-1)}=\frac{2x^2-x+1}{x^3-x^2+x-1}\]
The \(A,B,C\ldots\) are unknown coefficients that we want to find so that
\[\text{complicated rational function}=\text{sum of less complicated rational functions}\]

Answer 16

oooh I see, so now that the equation has been broken up, how would one find each coefficient since it equals to x^2? I vaguely understand the reverse process but im kind of lost on when it comes to finding each letter

Answer 17

on the equation p(x)/q(x) does that work to find every integral?

Answer 18

Here's a nice resource for review: http://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx

For this particular decomposition, it's a matter of solving a system of 5 equations with 5 unknowns, namely the one you get from matching up the coefficients of corresponding powers of \(x\).

For instance, if I had
\[x=(A+B)x^2+(B-A)x\]
then I get the system
\[\begin{cases}A+B=0\\B-A=1\end{cases}\]
For this function, you have
\[0x^4+0x^3+x^2+0x+0=(A+B)x^4+(-A+B+C)x^3+(A+C+D)x^2\\
\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+(-A+D+E)x+(A+E)\]
which gives the system,
\[\begin{cases}\begin{align*}A+B&=0\\-A+B+C&=0\\A+C+D&=1\\-A+D+E&=0\\A+E&=0\end{align*}\end{cases}\]

Answer 19

oooh, i find the link VERY helpful, do you mind if i go and try this on my own? Your explanation is great as well btw!

Answer 20

By all means. And you're welcome!