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[TheDarkStar]

What's the derivative and anti-derivative of e^x/2? It's not just e^x/2 is it?

The derivative of e^x/2 is (1/2)e^x/2 as TBF said.

The antiderivative of e^x/2 is also straightforward:

Taking u = x/2 and thus du = (1/2)dx -> 2du = dx,
∫e^x/2dx = ∫e^u * (2du)
= 2∫e^udu
= 2e^u + C
and substituting x/2 back in for u you get:
2e^x/2 + C

Integrating expressions involving exponentials ends up relying on getting the expression to equal some form of e^udu if you can't get rid of them with cancelling, expansions, substitution, integration by parts (you'll learn about this last one in Calculus 2/AP Calculus BC or maybe AB depending on the teacher/some equivalent course), etc.

[TheBiggerFish]

Hoo boy, advanced stuff.

[frostshotgg]

Oh right, U substitution. It's been so long since I've done basic derivatives/integrals without my TI-89 that I forgot what you actually do. I miss when math was easy and made sense.

[TheBiggerFish]

TI-89...
*drools*
They don't let you use them on the tests though...

[TheDarkStar]

Hoo boy, advanced stuff.

It's not too bad - the integration itself is usually covered in the first calculus class people take. In addition, it looks a lot better on paper than typed. However, it occurs to me that you might not have learned about u-substitution for solving integrals yet (but doing exponentials before u-substitution would be strange).

[TheBiggerFish]

We just haven't touched integrals yet.

[TheDarkStar]

We just haven't touched integrals yet.

Oh I see. That makes sense now.

For finding the antiderivative of e^x/2 without integration, you could look at it like this:

For an expression of the form b*e^ax, differentiating gets you b*(e^ax)(a) = a*b*e^ax. Since find the antiderivative undoes the derivative operation, the antiderivative of a*b*e^ax is just the original expression. In your case, a*b = 1, or with some rearranging b = 1/a (because the coefficient of e^x/2 is one). Rewriting, you get a*(1/a)*e^ax. Undoing the differentiation, the result is (1/a)*e^ax. Using your example where a = 1/2, you get 2e^x/2.

The concept can probably be worded better, but hopefully it gets at least some of the idea across.

Edit: Fixed.

[TheBiggerFish]

The example was a=1/2.  But, I sorta get it.

[TheDarkStar]

The example was a=1/2.

Fixed (and I had messed up in the last step anyway).

[Graknorke]

Is antiderivative a genuine term?

[Sacasco]

Is antiderivative a genuine term?

Yes, I think it is.
From what I understand, the indefinite integral is the set of all antiderivatives. That is, if f(x)=2x (let's keep things simple), the indefinite integral is x^2+c, but, for example, x^2+5 would be an antiderivative, as would x^2-6 and every other possible value of c.
Please, someone correct me if I'm wrong.

[MagmaMcFry]

Is antiderivative a genuine term?

Yes, I think it is.
From what I understand, the indefinite integral is the set of all antiderivatives. That is, if f(x)=2x (let's keep things simple), the indefinite integral is x^2+c, but, for example, x^2+5 would be an antiderivative, as would x^2-6 and every other possible value of c.
Please, someone correct me if I'm wrong.

Antiderivatives and indefinite integrals of f are the exact same thing, they're just preimages of f under the differentiation operator, i.e. functions whose derivative is f. The +c notation is just confusing and stupid if you don't explain what the c is supposed to do there.

[Spehss _]

What's the anti-derivative of lnx? (1/2)(lnx)²?

[Spehss _]

Um, no.

int(ln(x)dx)=x*ln(x)-int(dx)=x*ln(x)-x, I think.

As for how, it'd be integration by parts

I'm asking since it's a part of an integration by parts problem already. Guess I'll just use the derivative of lnx and the antiderivative of the other part instead, and see how that goes.

[Spehss _]

Well, integration by parts can be nested.

I have found that out already, yes. _{^{i hate it}}

When integrating by parts, how does one choose which part is used for the derivative and which part is used for the anti-derivative without trial and error? I keep picking the wrong sides for each problem and wind up having to erase everything and start a problem over, making homework take effectively twice as long. _{^{i hate it so much}}