# Why does using u-substitution give a different answer? [duplicate]

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?

• The first thing I would do is factor and kill those two minus signs... – Matthias Dec 15 '18 at 20:54

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 $$C = \frac 12 + c$$.

• 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). – amWhy Dec 15 '18 at 21:11

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.