# Integral Properties: Real Estate Solutions

• AquaGlass
In summary, the conversation discusses differentiation and integration as inverse operations. The integral of f'(x) dx is simply f(x). The integral of e^(tan^-1(x)) is e^(tan^-1(x)) + C, where C is the constant. The function should be evaluated over the interval [a,b], and there is no such thing as an indefinite integral, but it is commonly used as shorthand notation. The notation \int^x_a f(t) dt means finding any anti-derivative of f(x) with the constant a. Lastly, the integral should be evaluated over the interval [0,1] plus an arbitrary constant.
AquaGlass
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Last edited:
Differentiation and integration are inverse operations.

yep, the integral of f'(x)dx is simply f(x).

oh ok I see, so then it is just e ^ (tan^-1(x)) ?

+C, do not forget the constant, sometimes it can be really painful if you forget to add a constant at the end.

Oh ok, also do I evaluate that function over the interval [a,b] then? I forgot to mention it before.

AquaGlass said:
Oh ok, also do I evaluate that function over the interval [a,b] then? I forgot to mention it before.

Are u taking the indefinite or definite integral of that function? Why don't you show the original question first? it usually makes it easier for everyone!

Theres really no such thing as an indefinite integral =] Its just commonly used shorthand notation.

$$\int f(x) dx$$ really means $$\int^x_a f(t) dt$$ where a is some constant.

Gib Z said:
Theres really no such thing as an indefinite integral =] Its just commonly used shorthand notation.

$$\int f(x) dx$$ really means $$\int^x_a f(t) dt$$ where a is some constant.
No, it doesn't. $$\int f(x) dx$$ is any anti-derivative of f- it involves an arbitrary constant. The "a" in $$\int_a^x f(x) dx$$ determines a specific constant.

yep, just evaluate it over the interval [0,1]

$$\int^x_a f(t) dt$$ plus an arbitrary constant then =]

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