hestyllvql

2022-05-01

Antiderivative and definite integral

If f is a continuous, real-valued function on interval [a,b], then the fundamental theorem of calculus tells us that

${\int}_{a}^{x}f(t)dt=F(x)$

where F(x) is antiderivative, i.e. $F(x{)}^{\prime}=f(x)$.

If so, why I can't find the equality $\int f(x)dx={\int}_{a}^{x}f(t)dt=F(x)$ anywhere? It expresses the relationship between definite and indefinite integral in such a straightfoward way (assuming this equality is true). So it it true and can I use $\int $ and ${\int}_{a}^{x}$ interchangeably?

If f is a continuous, real-valued function on interval [a,b], then the fundamental theorem of calculus tells us that

${\int}_{a}^{x}f(t)dt=F(x)$

where F(x) is antiderivative, i.e. $F(x{)}^{\prime}=f(x)$.

If so, why I can't find the equality $\int f(x)dx={\int}_{a}^{x}f(t)dt=F(x)$ anywhere? It expresses the relationship between definite and indefinite integral in such a straightfoward way (assuming this equality is true). So it it true and can I use $\int $ and ${\int}_{a}^{x}$ interchangeably?

Zemmiq34

Beginner2022-05-02Added 11 answers

Step 1

Actually, the Fundamental Theorem of Calculus states that if $f:[a,b]\to \mathbb{R}$ is a continuous real-valued function, and F is its antiderivative, then

${\int}_{a}^{x}f(t)dt=F(x)-F(a)$

Step 2

This is precisely why, more accurately,

${\int}_{a}^{x}f(t)dt=F(x)-F(a)=\int f(x)dx+C$ for some $C\in \mathbb{R}$

Actually, the Fundamental Theorem of Calculus states that if $f:[a,b]\to \mathbb{R}$ is a continuous real-valued function, and F is its antiderivative, then

${\int}_{a}^{x}f(t)dt=F(x)-F(a)$

Step 2

This is precisely why, more accurately,

${\int}_{a}^{x}f(t)dt=F(x)-F(a)=\int f(x)dx+C$ for some $C\in \mathbb{R}$

RormFrure6h1

Beginner2022-05-03Added 13 answers

Step 1

You are probably referring to this part of the fundemental theorem of calculus

$\frac{d}{dx}{\int}_{a}^{x}f(t)dt=f(x)$

And your using the fact that $f(x)=\frac{d}{dx}\int f(x)dx$ which yields $\frac{d}{dx}{\int}_{a}^{x}f(t)dt=\frac{d}{dx}\int f(x)dx$.

Step 2

While this is true, remember that just because two functions have the same derivative, doesn't mean that they're the same function. Here is an example

$\begin{array}{rl}f(x)& ={\int}_{3}^{x}\mathrm{sin}tdt=-\mathrm{cos}x+\mathrm{cos}3\\ g(x)& =\int \mathrm{sin}xdx=-\mathrm{cos}x\end{array}$

As you can cleary see both functions have the same derivative, but the functions are not the same. Your statement will be true if a is a root of the indefinite integral. Example $\begin{array}{r}{\int}_{0}^{x}2tdt={x}^{2}-0={x}^{2}=\int 2xdx\end{array}$.

So it really depends on the function you are using. But the statement is NOT always true.

You are probably referring to this part of the fundemental theorem of calculus

$\frac{d}{dx}{\int}_{a}^{x}f(t)dt=f(x)$

And your using the fact that $f(x)=\frac{d}{dx}\int f(x)dx$ which yields $\frac{d}{dx}{\int}_{a}^{x}f(t)dt=\frac{d}{dx}\int f(x)dx$.

Step 2

While this is true, remember that just because two functions have the same derivative, doesn't mean that they're the same function. Here is an example

$\begin{array}{rl}f(x)& ={\int}_{3}^{x}\mathrm{sin}tdt=-\mathrm{cos}x+\mathrm{cos}3\\ g(x)& =\int \mathrm{sin}xdx=-\mathrm{cos}x\end{array}$

As you can cleary see both functions have the same derivative, but the functions are not the same. Your statement will be true if a is a root of the indefinite integral. Example $\begin{array}{r}{\int}_{0}^{x}2tdt={x}^{2}-0={x}^{2}=\int 2xdx\end{array}$.

So it really depends on the function you are using. But the statement is NOT always true.

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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