Question

Let y=∫ sin(t^4)dt on the interval (3x+1, 2x). Determine y '. I just don't know how to even begin. Thanks in advance!

Answer 1

Do you know about the fundamental theorem of Calculus?

Answer 2

Yes, I was just trying to apply that. :)

Answer 3

We might observe that the integral will end up having a form like this:
F(2x) - F(3x + 1)
where
dF(t)/dt = sin(t^4)
F(t) is the antiderivative of the integrand.

The derivative with respect to x of the integral is equivalent to taking the derivative of this form:
d/dx ( F(2x) - F(3x+1) )
where we just need to distribute and apply chain rule.

Answer 4

For the first term, as an example:
\( \dfrac{d}{dx} F(2x) = \dfrac{dF(2x)}{d(2x)} \times \dfrac{d(2x)}{dx} \)
\(= 2 f(2x) \)

f(x) is just our original integrand (but swapped t for x). f(x) = sin (x^4 )

Answer 5

If what I just said makes no sense, I can walk through it or maybe you were thinking of the derivative of an integral statement: \( \dfrac{d}{dx} \int \limits_{a}^{x} f(t) \ dt = f(x) \), I could try that route as well.

Answer 6

You lost me on the example. I have that formula you mentioned last written down as the first step in the FTC. If you could walk me through that way - that would be fantastic. :)

Answer 7

Because in that formula, we have just an x in the upper bound and a constant in the lower bound. Our integral looks like this:
\( \displaystyle \int_{3x+1}^{2x} \sin t^4 \ dt \)
First, we need to split up this integral so that one of the bounds is a constant.

Answer 8

So we can use this identity: \( \int_{a}^{b} = \int_{a}^{c} + \int_{c}^{b} \) (the same stuff is written for each integral, so it is easier to write this way) to at least put the constant in.
\( \displaystyle \int_{3x+1}^{2x} \sin t^4 \ dt = \int_{3x+1}^{c} \sin t^4 \ dt + \int_{c}^{2x} \sin t^4 \ dt \)
That part makes sense?

Answer 9

OK, I'm with you on that part. :)

Answer 10

After that, we are going to use the boundary flip identity: \(\int_{a}^{b} ...= \color{blue}-\int_{b}^{a}... \) for our first term, because the variable has to be on the top part.
\( \displaystyle \int_{3x+1}^{c} \sin t^4 \ dt = \color{blue}- \int_{c}^{3x + 1} \sin t^4 \ dt \)

So now we should have:
\( \displaystyle \int_{3x+1}^{2x} \sin t^4 \ dt = -\int_{c}^{3x+1} \sin t^4 \ dt + \int_{c}^{2x} \sin t^4 \ dt \)

Really these are all just integral properties that should be well-known. :)

Answer 11

Yes! Got it. That all looks familiar now. :) Thank you!!

Answer 12

Glad to help! The last step is taking the derivative of each side, and we will need to make sure to use chain rule to deal with the upper bound not being just x.
You might prefer to use a substitute variable: u = 3x + 1 and du/dx = 3.
\(\begin{align}
\dfrac{d}{dx} \int_{c}^{3x+1} \sin t^4 \ dt &= \dfrac{d}{dx} \int_{c}^{u} \sin t^4 \ dt \\ &= \dfrac{d}{du} \int_{c}^{u} \sin t^4 \ dt \times \dfrac{du}{dx} \\ &= \sin(u^4) \times \dfrac{du}{dx} \\ &= \sin ((3x+1)^4) \times 3
\end{align}\)

Answer 13

Fantastic!! Thank you so much!