Question

Fundamental Theorem of Calculus:
Find d/dx integral [1,x^4] sect dt

Answer 1

\[\large \int_1^{x^4}\sec dt = F(x^4)-F(1) \] now differentiate the right hand side of your equation, don't forget to use the chain rule and remember that\[\large F'(x)=f(x) \]

Answer 2

I'm really in sorry shape here - so F is supposed to be the antiderivative of sec right?

Answer 3

exactly, but if you differentiate it you will get \(f(x)\) again. Which is easy to work with, \(\sec(t)dt\) is not too straightforward to integrate, however applying the Fundamental Theorem is.

Notice that \(F(x^4)\) is the antiderivative evaluated at \(x^4\) (therefore another function) and \(F(1)\) is just a constant number.

Answer 4

ok so it should look like sec(x^4) - sec(1).
then applying the chain rule it would just be sec(x^4) (4x^3)

Answer 5

or actually shouldn't the sec change?

Answer 6

\[\large \frac{d}{dx}(F(x^4)-F(1))= F'(x^4) \cdot 4x^3 - 0 = 4x^3f(x) = 4x^3 \sec x \]

Answer 7

because \(f(t)= \sec (t) \)

Answer 8

pardon me, I messed up with the substitution above, let me copy and correct: \[\large \frac{d}{dx}(F(x^4)-F(1))= F'(x^4) \cdot 4x^3 - 0 = 4x^3f(x^4) = 4x^3 \sec (x^4) \tag{*} \]
where (*) is the correction, note that:
\[\large F'(x^4)=f(x^4) = \sec (x^4) \]

Sorry about that.

Answer 9

Ok, the original function is f. The antiderivative of f is F. And taking the derivative of F just takes me back to f.

Answer 10

And the last part I only needed to do because of the chain rule

Answer 11

you got it :-)

Answer 12

Thanks for your patience! :)

Answer 13

you're very welcome