What is the difference between the two parts of FTOC?

Sarai Davenport
2022-06-28
Answered

What is the difference between the two parts of FTOC?

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g2joey15

Answered 2022-06-29
Author has **21** answers

No, they are not the same. FTC1 is the big gun: It states that if $f$ is continuous on $[a,b],$, then f has an antiderivative $F,$, namely the function

$F(x)={\int}_{a}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\in [a,b].$Recall that before the student has even seen FTC1, a lot of work has already been done in guaranteeing that the integrals${\int}_{a}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$ even exist (limits of Riemann sums and all that). FTC1 is a crowning acheivement in that it says not only do those integrals exist, the derivative of the function so formed gives us back $f.$.

FTC2 is a lesser acheivement. All it says is that if you have any antiderivative $G$ of a continuous function $f$ on $[a,b],$, then ${\int}_{a}^{b}f(t)\phantom{\rule{thinmathspace}{0ex}}dt=G(b)-G(a).$. In FTC1 we already had an antiderivative, namely the $F$ defined there, which does this. FTC2 simply says any antiderivative will do this. The proof of FTC2 is almost trivial: By the MVT, any two antiderivatives on an interval differ by a constant, and the result follows.

$F(x)={\int}_{a}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\in [a,b].$Recall that before the student has even seen FTC1, a lot of work has already been done in guaranteeing that the integrals${\int}_{a}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$ even exist (limits of Riemann sums and all that). FTC1 is a crowning acheivement in that it says not only do those integrals exist, the derivative of the function so formed gives us back $f.$.

FTC2 is a lesser acheivement. All it says is that if you have any antiderivative $G$ of a continuous function $f$ on $[a,b],$, then ${\int}_{a}^{b}f(t)\phantom{\rule{thinmathspace}{0ex}}dt=G(b)-G(a).$. In FTC1 we already had an antiderivative, namely the $F$ defined there, which does this. FTC2 simply says any antiderivative will do this. The proof of FTC2 is almost trivial: By the MVT, any two antiderivatives on an interval differ by a constant, and the result follows.

asked 2022-06-15

Prove

${\int}_{-a}^{a}f(x)dx=0$

assuming $f(x)$ is odd.

${\int}_{-a}^{a}f(x)dx=0$

assuming $f(x)$ is odd.

asked 2022-05-13

Calculate:

${\int}_{-2}^{2}\mathrm{sin}({x}^{5}){e}^{({x}^{8}\mathrm{sin}({x}^{4}))}dx$

${\int}_{-2}^{2}\mathrm{sin}({x}^{5}){e}^{({x}^{8}\mathrm{sin}({x}^{4}))}dx$

asked 2022-06-06

Is FTOC valid only in interval that the function is continuous?

${\mathrm{\partial}}_{x}{\int}_{0}^{x}f({x}^{\prime})\phantom{\rule{thinmathspace}{0ex}}d{x}^{\prime}=f(x)-f(0)$

${\mathrm{\partial}}_{x}{\int}_{0}^{x}f({x}^{\prime})\phantom{\rule{thinmathspace}{0ex}}d{x}^{\prime}=f(x)-f(0)$

asked 2022-05-10

The FTOC states that if $f$ is continuous on $[a,b]$ then it is integrable.

If $f$ is not defined at certain points of $[a,b]$ we can often give meaning to an improper integral. But under what circumstances will $f$ always be integrable on it's domain, properly or improperly?

If we partition $[a,b]$ and consider $f$ on the subintervals $({x}_{i},{x}_{i+1})$ and assume it is continuous, (so $f$ is piecewise continuous on $[a,b]$), is $f$ then integrable on the entire interval simply by the adding up each integral over each subinterval? But the interval is open, so how does FTOC and improper integrals come into play? Will the improper integral over $[{x}_{i},{x}_{i-1}]$ always be well-defined?

If $f$ is not defined at certain points of $[a,b]$ we can often give meaning to an improper integral. But under what circumstances will $f$ always be integrable on it's domain, properly or improperly?

If we partition $[a,b]$ and consider $f$ on the subintervals $({x}_{i},{x}_{i+1})$ and assume it is continuous, (so $f$ is piecewise continuous on $[a,b]$), is $f$ then integrable on the entire interval simply by the adding up each integral over each subinterval? But the interval is open, so how does FTOC and improper integrals come into play? Will the improper integral over $[{x}_{i},{x}_{i-1}]$ always be well-defined?

asked 2022-07-01

How do you use the Fundamental Theorem of Calculus to find the derivative of $\int \frac{1}{1+{t}^{2}}dt$ from $x$ to $5$?

asked 2022-04-07

Let $f$ be an integrable function on $[a,b]$, $F$ be the antiderivative of $f$. Then

${\int}_{a}^{b}f\phantom{\rule{thinmathspace}{0ex}}du=F(b)-F(a)$

Is it suggesting that while $f$ is integrable, the derivative of the integral may not be $f$?

${\int}_{a}^{b}f\phantom{\rule{thinmathspace}{0ex}}du=F(b)-F(a)$

Is it suggesting that while $f$ is integrable, the derivative of the integral may not be $f$?

asked 2022-07-02

Finding $f(x)$ in ${\mathrm{cos}}^{2}(x)f(x)={x}^{2}-2{\int}_{1}^{x}\mathrm{sin}(t)\mathrm{cos}(t)f(t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t$