Question

help! calculus question

Answer 1

Finding a primitive function of \(f(x)=e^{t^3}\) is impossible. But, if we call that primitive function F, we know this:

\(\int_{a}^{b}f(t)dt=F(b)-F(a)\).

In this case, we get:

\(\dfrac{d}{dx}(F(x^2)-F(3))\).

Try to see what that leads to...

Answer 2

how would i solve the F(x^2)? substitute that in as the t values?

Answer 3

put x^2 = w, and use fundamental theorem of calculus

Answer 4

and chain rule

Answer 5

\(\dfrac{d}{dx}(F(x^2)-F(3)) \)


say, \(x^2 = w\)

\(\dfrac{d}{dx}(F(w)) \)


\(f(w) \dfrac{d}{dx}(w) \)


\(e^{w^3} \dfrac{d}{dx}(w) \)

Answer 6

Once you have \(F(x^2)\), the t is gone!
Also, F(3) is a constant, so its derivative is 0.

Now, \(F(x^2)\) is a chain function (of x), so you have to apply the Chain Rule:
I use this notation, which I find more clear than the d\dx one:

\((F(x^2))'=F'(x^2)\cdot(x^2)'=e^{(x^2)^3}\cdot2x=2x e^{x^6}\)

Answer 7

Thank you! It all makes sense now!

Answer 8

YW!