Gretchen Schwartz

2022-07-08

So this has been confusing me a lot. Let $f\left(x\right)={x}^{2}$ and let $F\left(x\right)={\int }_{1}^{x}f\left(x\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$. Then $F\left(1\right)=0$, obviously, but the antiderivative of f (which is the same as ${\int }_{1}^{x}f\left(x\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$) is $F\left(x\right)={x}^{3}/3$, so $F\left(1\right)=1/3$, not 0?
What caused the confusion is that I forgot that a function has infinitely many anti derivatives; they are defined in this case as $F\left(x\right)={x}^{3}/3+C$, and in this example $F\left(x\right)={x}^{3}/3-1/3$, ${x}^{3}/3$ is just a special antiderivative where $C=0$

treccinair

Expert

Explanation:
It does not matter what C is equal to.
The antiderivative of f is
$F\left(x\right)={\int }_{0}^{x}f\left(t\right)dt={\int }_{0}^{x}{t}^{2}dt=\left(\frac{{t}^{3}}{3}+C\right){|}_{0}^{x}=\left(\frac{{x}^{3}}{3}+C\right)-\left(\frac{{0}^{3}}{3}+C\right)=\frac{{x}^{3}}{3}$
$\therefore F\left(1\right)={\int }_{0}^{1}f\left(t\right)dt=1/3$

therightwomanwf

Expert

Step 1
The integral ${\int }_{1}^{x}{x}^{2}\phantom{\rule{thinmathspace}{0ex}}dx$ is equal to
${\frac{{x}^{3}}{3}|}_{1}^{x}.$
This is $\frac{1}{3}{x}^{3}-\frac{1}{3}$. Note that $\frac{1}{3}{x}^{3}-\frac{1}{3}$ is 0 at $x=1$, as you pointed out it should be.
Step 2
Remark: When we are finding definite integrals with limits of integration that involve variables, it is best to use a different "dummy" variable of integration. Things may be clearer if you write
$F\left(x\right)={\int }_{1}^{x}{t}^{2}\phantom{\rule{thinmathspace}{0ex}}dt.$
While in principle the notation you used is not wrong, it is, for good reason, frowned on.
Note also that talking about "the" antiderivative of a function f(t) is not quite right. The function has infinitely many (very closely related) antiderivatives.

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