Integral(e^(ab))da= e^(ab)*b^-1

Why?

Question

Integral(e^(ab))da=  e^(ab)*b^-1

Why?

anonymous · Answer

As, going the other way: if I differentiate the right equation with respect to a, b doesn't dissapear, as it is a constant, then why does a b^-1 materialise in the antiderivative?

amistre64 · Answer

i take it thats a partial and not b(a)?

amistre64 · Answer

$$D_a(exp({ab}))=exp(ab)*(ab)'$$
$$D_a(exp({ab}))=exp(ab)(a'b+ab')$$
or maybe if bs a constant

$$D_a(exp({ab}))=exp(ab)*b$$

amistre64 · Answer

we need the 1/b to catch it; 1/b * b = 1

amistre64 · Answer

in other words, if we dont put something there to catch the "b" that flies out we dont get the derivative we are looking for

amistre64 · Answer

Dx is a way to notate a derivative operation on something.

amistre64 · Answer

Da means take the derivative of this stuff with respect to a

amistre64 · Answer

go ahead and replace b with your favorite constant thats not 0

amistre64 · Answer

youll see that if we dont have a 1/2 in the antiderivative that we end up with :

   exp(2a)*2  which is NOT what we are trying to undo.

amistre64 · Answer

so we apply a useful form of 1 into the antiderivative to help us out; since 1* anything doesnt change its value;  we need a "2", or a "b" as the case may be, so lets use 2/2 and just pull out the 1/2 for later

anonymous · Answer

Thanks, after writing that out it clicked.

amistre64 · Answer

:)  youre welcome