When I try to integrate $(1-x)^2$ I get $\dfrac13(1 - x)^3$. However, when I try this integral on an online integration portal they all give me $\dfrac13(x - 1)^3$ instead. Can someone please explain why my computation is wrong?
What confuses me here is the negation. Creating the first derivative from the result would produce $(x-1)^2$, which is not the same as what I began with.
On
Substitute $u=1-x$ so $du=-dx$ and$$\int(1-x)^2dx=-\int u^2du=-\frac13(1-x)^3+C=\frac13(x-1)^3+C.$$Or if you prefer, substitute $v=x-1$ so $dv=dx$ and$$\int(1-x)^2dx=\int(x-1)^2dx=\int v^2dv=\frac13v^3+C=\frac13(x-1)^3+C.$$More generally, if $w=ax+b$ then $dw=adx$ and $dx=\frac{dw}{a}$, so$$\int(ax+b)^2dx=\int\frac1aw^2dw=\frac{1}{3a}w^3+C=\frac{1}{3a}(ax+b)^3+C.$$
On
If $F$ is an antiderivative of $f$ and $a\neq0$, then an antiderivative of $f(ax)$ will be $\frac1aF(ax)$, and not simply $F(ax)$ (if you differentiate this last function, what you get is $aF(ax)$).
In particular (taking $a=-1$), an antiderivative of $f(-x)$ is $\frac1{-1}F(-x)=-F(-x)$.
Now, apply this to $f(x)=(1+x)^2$ and $F(x)=\frac13(1+x)^3$.
On
What confuses me here is the negation. Creating the first derivative from the result would produce $(x-1)^2$, which is not the same as what I began with.
Actually, it is the same. Remember that the square of an expression is the same as the square of its negative: $a^2 \equiv (-a)^2$. So $(x-1)^2 \equiv (1-x)^2$.
If you differentiate your original answer, $\frac{1}{3}(1-x)^3$, you'll get $-(1-x)^2$. This is because of the chain rule; you differentiate the expression as if with respect to the inner function $1 - x$, then multiply by the derivative of this inner function, $-1$.
However, we can use this 'guess' to find the true answer. Write what we know:
$$\frac{d}{dx}(\frac{1}{3}(1-x)^3) = -(1-x)^2 $$
Applying linearity of derivatives we can move the negative to the other side:
$$\frac{d}{dx}(-\frac{1}{3}(1-x)^3) = (1-x)^2$$
Which of course implies that:
$$\int (1-x)^2 \, dx = -\frac{1}{3}(1-x)^3 +C = \frac{1}{3}(x-1)^3 + C$$
If you can think about integration like this in your head, it's a lot faster than $u$-substitution for simple functions.
On integrating $(1-x)^2$ you get,
$$\int{(1-x)^2 dx} = - \frac{1}{3} (1-x)^3 $$
Which is equal to $\frac{1}{3} (x-1)^3$.
You missed the negative sign after integrating your expression!
Hope it helped!