- #1

- 145

- 9

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)

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- Thread starter Saracen Rue
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- #1

- 145

- 9

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

- #2

- 3,987

- 1,538

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

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

- 145

- 9

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

- 3,987

- 1,538

Yes, that should work.

- #5

- 145

- 9

Thank you! :)Yes, that should work.

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