1. ## Definite Integral close??

Find the derivative of

h(x) = integral(sin(x) at the top and -5 at the bottom) of
cos(t^5) +t

So... here's what I thought the answer should be, but I'm evidently missing something, again.

My plan of attack is to insert sin(x) for t and then subtract the function with -5 inserted for t:

cos(sin^5(x)) + sin(x) - (cos((-5)^5) - 5)

I'm getting told this is incorrect. Any glaring mistakes? I've been doing this for several hours, so I may have gone mentally blind.

Thanks!

2. $h'(x) = \frac{d}{dx} \int_{-5} ^{\sin x}\left( \cos t^5 + t\right)dt$

Main thing: Chain rule

So as a 'modified' version of the fundamental theorem of calculus: Let $F(x) = \int_a^{u(x)}f(t) \ dt$. Then $F'(x) = f \bigg( u(x) \bigg) u'(x)$ .

Imagine $f(t) = \cos t^5 + t$ and $u(x) = \sin x$.

Apply the fundamental theorem of calculus as you would normally to get: $h'(x) = \underbrace{\left[\cos \left(\sin^5 x\right) + \sin x\right]}_{f \big( u(x) \big)} \cdot \underbrace{\left(\sin x\right)'}_{u'(x)}$

Take the derivative of $\sin x$ and you're done.

3. So why didn't I have to take the -5 into consideration?

4. Well look at the fundamental theorem of calculus: $\frac{d}{dx} \int_{a}^{u(x)} f(t) \ dt = f \bigg( u(x) \bigg) u'(x)$ ... or ... $\frac{d}{dx} \int_{a}^{x} f(t) \ dt = f (x)$

In either case, you don't see the bottom limit of the integral, $a$, in the right hand side of the equation at all.

5. Okay, I see what you mean, but in earlier problems I've been finding the answer by plugging in the upper limit and subtracting the lower limit. This had been giving me the correct answer until this question.

For example, and earlier question said to find the derivative of upper limit = 1 lower = x
sin(t^3) The correct answer ended up being sin(1^3) - sin(x^3) or just -sin(x)

If I had followed the FTC and just used the upper limit I would've just gotten sin(1^3) or 0. How do I know when to do which?

6. FTC says: $\frac{d}{dx} \int_{a}^{x} f(t) \ dt = f (x)$ with x as the upper limit.

If it is on the bottom, you'll have to make the adjustment by using the fact that: $\int_a^b f(x) dx = -\int_b^a f(x) dx$

So for your question, by FTC: $\frac{d}{dx} \int_{x}^{1} \sin t^3 \ dt = -\frac{d}{dx} \int_1^x \sin t^3 \ dt = -\sin x^3$

7. Thank you!