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I am working on a problem that required finding $\int (-7-y)^2dy$. Instead of foiling it out, I just used u-substitution to find the antiderivative. ($u=-7-y; u'=-1$). I got an antiderivative of $\frac{-(-7-y)^3}{3}$. However, if you factor out this antiderivative you end up with a constant of $\frac {343}{3}$ that's not there if you foil then integrate. Why would the +c be defined for u-sub but not for foiling then integrating?

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    $\begingroup$ The first thing I would do is factor and kill those two minus signs... $\endgroup$ – Matthias Dec 15 '18 at 20:54
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Remember: Indefinite integration yields a family of functions that differ by only a constant.

$$\int f(x) \,dx = F(x) + C$$

What $C$ happens to be is unknown to us; we just know that if we differentiate $F(x)$, we get $f(x).$


Let's see how this works in a simpler example:

Suppose we have $$\int (x-1)\, dx$$

Method 1: Let $u = x-1$. Then $du = dx.$, and we have $$\int u\,du = \frac{u^2}2 + C= \frac{(x-1)^2}2 + c = \frac 12(x^2-2x + 1) + c = \frac 12 x^2 -x + \left(\frac 12 + c\right) \tag 1$$

Method 2: integrate directly: $$\int (x-1)\, dx = \frac{x^2}2 - x + C\tag 2$$

$(1)$ and $(2)$ differs only by a constant: We can make them "equal" by setting $C = \frac 12 + c$.

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  • $\begingroup$ The constant $\dfrac {343}3$ is not the "constant of integration" from the u-substitution method you used. The constant of integration from that method is $\dfrac{343}3 + c = C$, where $C$ is a place holder, which upon differentiation, disappears (and is the integrand). $\endgroup$ – amWhy Dec 15 '18 at 21:11
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Notice that with $u$ substitution you get $$\frac{-(-7-y)^3}{3}+C=\frac{y^3}{3}+7y^2+49y+\frac{343}{3}+C$$ (don't forget the $+C$). If you foil it then integrate you would get $$\int (y^2+14y+49)dy=\frac{y^3}{3}+7y^2+49y+C'$$ where $C'$ is some constant. Notice that these two anti-derivates still only vary by a constant, namely $\frac{343}{3}+C-C'$. Should you wish, you could simply relabel $\frac{343}{3}+C$ as $C''$ and the two answers would look symbolically the same.

At the end of the day we still have $$\left\{\frac{-(-7-y)^3}{3}+C\; :\; C\in\mathbb{R}\right\}=\left\{\frac{y^3}{3}+7y^2+49y+C'\; :\; C'\in\mathbb{R}\right\}$$ so these two ways of writing it give us the same set of solutions.

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