- #1

- 1,462

- 44

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Mr Davis 97
- Start date

- #1

- 1,462

- 44

- #2

Mark44

Mentor

- 36,019

- 7,948

The antiderivative of the derivative of a function is the function (plus an arbitrary constant). It's as simple as that.

In symbols: ##\int f'(t) dt = f(t) + C##

As a definite integral, along the lines of the equation you showed,

##\int_0^t f'(x)dx = f(t) - f(0)##. Here -f(0) is the constant C I showed above.

- #3

Stephen Tashi

Science Advisor

- 7,739

- 1,525

I am looking at a solution to an integral

What is the integral whose solution you seek?

- #4

- 1,462

- 44

Sorry, I made a mistake. The first line should be ##\displaystyle f(t) = \int_0^1 \frac{\log (tx+1)}{x^2+1} ~ dx##. The integral I seek is ##f(1)##.What is the integral whose solution you seek?

- #5

- 1,462

- 44

I guess the use of ##t## is a limit of integration is confusing me, since t is already a variable elsewhere. For example, say that ##\frac{dy}{dx} = 2x##. Then ##\int_0^x dy = \int_0^x 2x ~dx##. Doesn't this latter statement seem kind of weird? Why is x a limit of integration if it is already used as a variable? Also, what does ##\int_0^x dy## turn out to look like?The antiderivative of the derivative of a function is the function (plus an arbitrary constant). It's as simple as that.

In symbols: ##\int f'(t) dt = f(t) + C##

As a definite integral, along the lines of the equation you showed,

##\int_0^t f'(x)dx = f(t) - f(0)##. Here -f(0) is the constant C I showed above.

- #6

Mark44

Mentor

- 36,019

- 7,948

A limit of integration shouldn't be used also as a variable in the integrand (AKA "dummy variable"). You should note that in my 2nd example I was careful to use different variables for the two different roles.I guess the use of ##t## is a limit of integration is confusing me, since t is already a variable elsewhere. For example, say that ##\frac{dy}{dx} = 2x##. Then ##\int_0^x dy = \int_0^x 2x ~dx##. Doesn't this latter statement seem kind of weird? Why is x a limit of integration if it is already used as a variable? Also, what does ##\int_0^x dy## turn out to look like?

If ##\frac{dy}{dx} = 2x##, then ##dy = 2x dx##, so ##\int_0^t dy = \int_0^t 2x dx \Rightarrow y(t) - y(0) = x^2|_0^t \Rightarrow y(t) = t^2 + y(0)##

- #7

- 1,462

- 44

Okay, so the fact the solution I am reading does use the same variable as the variable in the integrand is not technically wrong, but is just bad "style"?A limit of integration shouldn't be used also as a variable in the integrand (AKA "dummy variable"). You should note that in my 2nd example I was careful to use different variables for the two different roles.

If ##\frac{dy}{dx} = 2x##, then ##dy = 2x dx##, so ##\int_0^t dy = \int_0^t 2x dx \Rightarrow y(t) - y(0) = x^2|_0^t \Rightarrow y(t) = t^2 + y(0)##

- #8

Mark44

Mentor

- 36,019

- 7,948

Yes, and confusing to some.Okay, so the fact the solution I am reading does use the same variable as the variable in the integrand is not technically wrong, but is just bad "style"?

Share:

- Replies
- 2

- Views
- 6K