Difficult 2u integration of exponentials - HELP please (1 Viewer)

[kaz1]

reverse chain rule, pretty much substitution with first principles, remember it in the cambridge book back in the day

try to make the outside polynomial bit the derivative of the exponent

 

[zeebobDD]

differentiate y=e^x^(3) -1 dy/dx = e^x^(3)-1 * 3x^2

therefore integrate both sides therefore, e^x^(3) -1 / 3

you move the 3 down from the derivative

 

[Nooblet94]

Any chance this is a trapezoidal/Simpson's rule question?

Nope, it's actually a pretty simple integral once you know what you're doing - it just looks really nasty

reverse chain rule, pretty much substitution with first principles, remember it in the cambridge book back in the day

try to make the outside polynomial bit the derivative of the exponent

That's how you do it.

 

[Shadowdude]

My apologies. I have no idea how to approach the question then.

p.s how do you start a new line in LATEX?

\\

or Just go to a new line here and use the 'tex' tags again.

 

[Sanjeet]

d/dx e^f(x) = f'(x) e^f(x)

So integral f'(x) e^f(x) dx = e^f(x) + c

So you recognise the "pattern" in 2U.

Nope. that wasn't in the form of f'(x)e^f(x)

 

[Carrotsticks]

Nope. that wasn't in the form of f'(x)e^f(x)

Yes it is.

 

[Sanjeet]

Yes it is.

is in the form of f'(x)e^f(x) ????? Am I blind or something?

 

[Shadowdude]

is in the form of f'(x)e^f(x) ????? Am I blind or something?

That's not, but the actual question:



is, in a way.

 

[Carrotsticks]

is in the form of f'(x)e^f(x) ????? Am I blind or something?

I am confused. Isn't the original question:

 

[Sanjeet]

I am confused. Isn't the original question:

yes but if you look at my original post you'll see what I was referring to lol

 

[Sy123]

Either way, it was silly of the question not to state the usual first part:
i) Find d/dx e^(x^3)
ii) Hence find integral

And even then, it isnt hard to see the relation, but yeah I would think its not exam suitable