Recent questions in Second Fundamental Theorem of Calculus

Integral CalculusAnswered question

julioloupiasvfx 2023-02-06

Explain the meaning of derived unit with the help of one example.

Integral CalculusAnswered question

lamesa1Vy 2022-12-04

Which of the following is TRUE ?

A

The fundamental period of $\mid \mathrm{sin}x\mid +\mid \mathrm{cos}x\mid \phantom{\rule{0ex}{0ex}}$

B

The fundamental period of $\mid \mathrm{sin}x\times \mathrm{cos}x\mid \phantom{\rule{0ex}{0ex}}$

C

The fundamental period of $\mid \mathrm{sin}x\mid -\mid \mathrm{cos}x\mid \phantom{\rule{0ex}{0ex}}$

D

The fundamental period of $\mid \mathrm{sin}x\mid $

A

The fundamental period of $\mid \mathrm{sin}x\mid +\mid \mathrm{cos}x\mid \phantom{\rule{0ex}{0ex}}$

B

The fundamental period of $\mid \mathrm{sin}x\times \mathrm{cos}x\mid \phantom{\rule{0ex}{0ex}}$

C

The fundamental period of $\mid \mathrm{sin}x\mid -\mid \mathrm{cos}x\mid \phantom{\rule{0ex}{0ex}}$

D

The fundamental period of $\mid \mathrm{sin}x\mid $

Integral CalculusAnswered question

Laila Murphy 2022-11-02

Second Fundamental theorem of calculus

Theorem 3.20 (Second Foundamental Theorem of Calculus)

Let f be a continuous function on [a,b] and F any function on [a,b], differentiable on (a,b), continuous on [a,b] such that ${F}^{\prime}(x)=f(x)$ for all $x\in (a,b)$.

Then

${\int}_{a}^{b}f(x)\mathrm{d}x=F(b)-F(a)$

I need to use the second Fundamental theorem of calculus to work out:

${\int}_{0}^{\frac{\pi}{8}}\mathrm{tan}(2x)\mathrm{d}x$

Firstly it is clear that tan(2x) is continuous on $[0,\frac{\pi}{8}]$

Now $F(x)=-\frac{1}{2}\mathrm{ln}|\mathrm{cos}(2x)|=-\frac{1}{2}\mathrm{ln}\mathrm{cos}(2x)$

where${F}^{\prime}(x)=f(x)$

To show that F(x) is differentiable $\mathrm{\forall}x\in (0,1)$ is it enough to say that as f(x) is continuous on (0,1) the derivative exists?

Theorem 3.20 (Second Foundamental Theorem of Calculus)

Let f be a continuous function on [a,b] and F any function on [a,b], differentiable on (a,b), continuous on [a,b] such that ${F}^{\prime}(x)=f(x)$ for all $x\in (a,b)$.

Then

${\int}_{a}^{b}f(x)\mathrm{d}x=F(b)-F(a)$

I need to use the second Fundamental theorem of calculus to work out:

${\int}_{0}^{\frac{\pi}{8}}\mathrm{tan}(2x)\mathrm{d}x$

Firstly it is clear that tan(2x) is continuous on $[0,\frac{\pi}{8}]$

Now $F(x)=-\frac{1}{2}\mathrm{ln}|\mathrm{cos}(2x)|=-\frac{1}{2}\mathrm{ln}\mathrm{cos}(2x)$

where${F}^{\prime}(x)=f(x)$

To show that F(x) is differentiable $\mathrm{\forall}x\in (0,1)$ is it enough to say that as f(x) is continuous on (0,1) the derivative exists?

Integral CalculusAnswered question

tonan6e 2022-10-02

Difference between first and second fundamental theorem of calculus

In first fundamental theorem of calculus,it states if $A(x)={\int}_{a}^{x}f(t)dt$ then ${A}^{\prime}(x)=f(x)$.But in second they say ${\int}_{a}^{b}f(t)dt=F(b)-F(a)$,But if we put x=b in the first one we get A(b).Then what is the difference between these two and how do we prove A(b)=F(b)−F(a)?

In first fundamental theorem of calculus,it states if $A(x)={\int}_{a}^{x}f(t)dt$ then ${A}^{\prime}(x)=f(x)$.But in second they say ${\int}_{a}^{b}f(t)dt=F(b)-F(a)$,But if we put x=b in the first one we get A(b).Then what is the difference between these two and how do we prove A(b)=F(b)−F(a)?

Integral CalculusAnswered question

aurelegena 2022-09-29

The notation of the second fundamental theorem of Calculus

I am self studying calculus, and just finished the lesson on the second fundamental theorem of calculus.

the way the theorem is described is:

$\frac{d}{dx}({\int}_{a}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt)=f(x)$

and it was told that the meaning is that the derivative of an integral of a function is the function itself.

I don't get how you can get that from this. the expression that I would think suggests this is:

$\frac{d}{dx}(\int f(x)\phantom{\rule{thinmathspace}{0ex}}dt)=f(x)$

so the derivative of an indefinite integral (as oppose to integrating over a range) of a function is the function itself.

another interpretation of the FToC2 I read here, is that it means that the derivative of the functions that gives the area under the curve of a different function is the different function. this is also something I don't understand how the FToC2 suggests of?

to me, it seems like what this means:

$\frac{d}{dx}({\int}_{a}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt)=f(x)$

is how a very small change in x affects that area under f(t) between a (a constant) and x. how do I get from that to the right interpretation?

I am self studying calculus, and just finished the lesson on the second fundamental theorem of calculus.

the way the theorem is described is:

$\frac{d}{dx}({\int}_{a}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt)=f(x)$

and it was told that the meaning is that the derivative of an integral of a function is the function itself.

I don't get how you can get that from this. the expression that I would think suggests this is:

$\frac{d}{dx}(\int f(x)\phantom{\rule{thinmathspace}{0ex}}dt)=f(x)$

so the derivative of an indefinite integral (as oppose to integrating over a range) of a function is the function itself.

another interpretation of the FToC2 I read here, is that it means that the derivative of the functions that gives the area under the curve of a different function is the different function. this is also something I don't understand how the FToC2 suggests of?

to me, it seems like what this means:

$\frac{d}{dx}({\int}_{a}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt)=f(x)$

is how a very small change in x affects that area under f(t) between a (a constant) and x. how do I get from that to the right interpretation?

Integral CalculusOpen question

Celeb G.2022-09-23

- ${\int}_{0}^{\mathrm{\pi}/4}\mathrm{sin}\left(4t\right)dt$
- ${\int}_{1}^{1}{x}^{n}dxwherenisapositiveeveninteger.$
- ${\int}_{0}^{\frac{\pi}{6}}2{e}^{\mathrm{sin}\left(x\right)}\mathrm{cos}\left(x\right)dx$
- $\int \frac{dx}{{(6x+5)}^{4}}$

Integral CalculusAnswered question

Celeb G.2022-09-23

- ${\int}_{0}^{\mathrm{\pi}/4}\mathrm{sin}\left(4t\right)dt$
- ${\int}_{-1}^{1}{x}^{n}dx$

Integral CalculusAnswered question

Andreasihf 2022-09-03

Second Fundamental Theorem of Calculus...

Let f have a continuous second derivative. Prove that

$f(x)=f(a)+(x-a){f}^{\prime}(a)+{\int}_{a}^{x}(x-t){f}^{\u2033}(t)dt.$

Here is my attempt at the problem.

Since f has a continuous second derivative, then the first derivative is also continuous. Therefore, by the first fundamental theorem of calculus, we have that

$f(x)=f(a)+{\int}_{a}^{x}{f}^{\prime}(t)dt.$

Expanding out the right-hand side of the above using integration by parts, we see that

$f(x)=f(a)+{f}^{\prime}(t)t-{\int}_{a}^{x}t{f}^{\u2033}(t)dt.$

This is where I am confused.

Let f have a continuous second derivative. Prove that

$f(x)=f(a)+(x-a){f}^{\prime}(a)+{\int}_{a}^{x}(x-t){f}^{\u2033}(t)dt.$

Here is my attempt at the problem.

Since f has a continuous second derivative, then the first derivative is also continuous. Therefore, by the first fundamental theorem of calculus, we have that

$f(x)=f(a)+{\int}_{a}^{x}{f}^{\prime}(t)dt.$

Expanding out the right-hand side of the above using integration by parts, we see that

$f(x)=f(a)+{f}^{\prime}(t)t-{\int}_{a}^{x}t{f}^{\u2033}(t)dt.$

This is where I am confused.

Integral CalculusAnswered question

Benedict Kokeh2022-08-19

Give Arrhenius and Bronsted definition of acid and base. Why are Bronsted's definition more useful in describing acid base properties

Integral CalculusAnswered question

Carsen Patel 2022-08-11

fundamental theorem of calculus and second derivative

Using the fundamental theorem of calculus, find the second derivative of

${\int}_{\sqrt{(}x)}^{x}x-{e}^{t}\phantom{\rule{thinmathspace}{0ex}}dt$

I've looked up the theorem on wikipedia but I can't really see what I'm meant to do.

Using the fundamental theorem of calculus, find the second derivative of

${\int}_{\sqrt{(}x)}^{x}x-{e}^{t}\phantom{\rule{thinmathspace}{0ex}}dt$

I've looked up the theorem on wikipedia but I can't really see what I'm meant to do.

The second fundamental theorem of calculus is a powerful tool for solving problems involving integration. It states that if a function is the derivative of another function, then the original function can be found by integrating the derivative. This theorem is useful in providing answers to complicated integrations. It is used widely in mathematics to solve equations, and to help students learn the fundamentals of calculus. Having a good understanding of the second fundamental theorem of calculus is essential for mastering the subject.