Gretchen Schwartz

Answered

2022-07-08

Simple question about antiderivative?

So this has been confusing me a lot. Let $f(x)={x}^{2}$ and let $F(x)={\displaystyle {\int}_{1}^{x}f(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x}$. Then $F(1)=0$, obviously, but the antiderivative of f (which is the same as ${\int}_{1}^{x}f(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$) is $F(x)={x}^{3}/3$, so $F(1)=1/3$, 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 $F(x)={x}^{3}/3+C$, and in this example $F(x)={x}^{3}/3-1/3$, ${x}^{3}/3$ is just a special antiderivative where $C=0$

So this has been confusing me a lot. Let $f(x)={x}^{2}$ and let $F(x)={\displaystyle {\int}_{1}^{x}f(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x}$. Then $F(1)=0$, obviously, but the antiderivative of f (which is the same as ${\int}_{1}^{x}f(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$) is $F(x)={x}^{3}/3$, so $F(1)=1/3$, 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 $F(x)={x}^{3}/3+C$, and in this example $F(x)={x}^{3}/3-1/3$, ${x}^{3}/3$ is just a special antiderivative where $C=0$

Answer & Explanation

treccinair

Expert

2022-07-09Added 18 answers

Explanation:

It does not matter what C is equal to.

The antiderivative of f is

$F(x)={\int}_{0}^{x}f(t)dt={\int}_{0}^{x}{t}^{2}dt=(\frac{{t}^{3}}{3}+C){|}_{0}^{x}=(\frac{{x}^{3}}{3}+C)-(\frac{{0}^{3}}{3}+C)=\frac{{x}^{3}}{3}$

$\therefore F(1)={\int}_{0}^{1}f(t)dt=1/3$

It does not matter what C is equal to.

The antiderivative of f is

$F(x)={\int}_{0}^{x}f(t)dt={\int}_{0}^{x}{t}^{2}dt=(\frac{{t}^{3}}{3}+C){|}_{0}^{x}=(\frac{{x}^{3}}{3}+C)-(\frac{{0}^{3}}{3}+C)=\frac{{x}^{3}}{3}$

$\therefore F(1)={\int}_{0}^{1}f(t)dt=1/3$

therightwomanwf

Expert

2022-07-10Added 4 answers

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

$F(x)={\int}_{1}^{x}{t}^{2}\phantom{\rule{thinmathspace}{0ex}}dt.$

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.

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

$F(x)={\int}_{1}^{x}{t}^{2}\phantom{\rule{thinmathspace}{0ex}}dt.$

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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