- #1

- 144

- 7

I know that ∫(a,b,f(x)dx = F(a) - F(b), so I was wondering if it's possible to define

*a*and

*b*so that the resulting definite integral equals the indefinite integral (where c = 0)

- B
- Thread starter Saracen Rue
- Start date

- #1

- 144

- 7

I know that ∫(a,b,f(x)dx = F(a) - F(b), so I was wondering if it's possible to define

- #2

- 3,886

- 1,454

$$F(x)=\int_a^x f(t)dt$$

then the sketch will be exactly the same as that of an indefinite integral ##G=\int f(x)dx## with zero integration constant, except shifted downwards by ##G(a)##.

- #3

- 144

- 7

So if I make

$$F(x)=\int_a^x f(t)dt$$

then the sketch will be exactly the same as that of an indefinite integral ##G=\int f(x)dx## with zero integration constant, except shifted downwards by ##G(a)##.

$$F(x)=\int_a^x f(t)dt$$

With no shift?

- #4

- 3,886

- 1,454

Yes, that should work.

- #5

- 144

- 7

Thank you! :)Yes, that should work.

- Replies
- 9

- Views
- 962

- Replies
- 11

- Views
- 1K

- Last Post

- Replies
- 10

- Views
- 955

- Replies
- 12

- Views
- 6K

- Replies
- 3

- Views
- 1K

- Replies
- 11

- Views
- 28K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 6

- Views
- 863

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 1K