Question

Use part I of the Fundamental Theorem of Calculus to find the derivative of
h(x) = \int_{-5}^{\sin(x)} (\cos(t^4)+t) dt

Answer 1

That would be... \(\int_{a(t)}^{b(t)}f(t)\;dt = f(b(t)) - a(b(t))\)

THAT "Fundamental Theorem"?

Answer 2

I believe so but I thought that was the formula for finding the antiderivative and this problem asks for the derivative

Answer 3

what is the derivative of F(b(t)) ?

Answer 4

and that was not the formula for finding an antiderivative ....

Answer 5

sinx*cos(x^4)+1?

Answer 6

more generally

the derivative of F(b(x)) since our limits are functions of x, is an application of the chain rule: F'(b(x)) * b'(x)

Answer 7

\[\int_{a(x)}^{b(x)}f(t)~dt=F(b(x))-F(a(x))\]
\[\frac d{dx}\int_{a(x)}^{b(x)}f(t)~dt=\frac d{dx}F(b(x))-\frac d{dx}F(a(x))\]

Answer 8

the derivative of F is already stated in the integral ... as f(t)

Answer 9

does this make sense?

Answer 10

so the derivative of F is cos(x^4)+x?

Answer 11

F'(b(x)) = f(b(x)) = cos((b(x))^4) + b(x)

Answer 12

the chain rule then multiplies that by the derivative of b(x)

Answer 13

cos*x(cos((sin(x))^4) + sin(x))?

Answer 14

cosx*(cos((sin(x))^4) + sin(x)) Is What I meant

Answer 15

yep

Answer 16

and F(a(x)) = F(-5)

and the derivative of a(x) = 0 so that term 'goes away'

Answer 17

We PLAY LIKE we can find the anti-derivative and then we use it. This is an awesome result.

Answer 18

Whoops. Just noticed type in first response. f(b(t)) not a(b(t)) Of course, amistre64 got it right all along.

Answer 19

other than an exercise in futility :) i do not see any practical application of taking the derivative of the integral

Answer 20

So when ever I have a problem like this I will want to first plug in b(x) for the variable and then if b(x) is something more complex then a single variable I will multiply that new equation by the derivative of b(x). Is this correct?

Answer 21

if that is how you picture the chain rule in action ... yes

Answer 22

you will want to apply both b(x) and a(x), both limits.


it just so happened in this case that a(x) = -5, so the derivative of a=0

F'(a(x)) * a'(x)

f(-5) * 0 = 0

Answer 23

and then subtract F'(b(x))*b'(x) from F'(a(x))*a'(x)?

Answer 24

other way about
\[\frac d{dx}[F(b)-F(a)]=f(b)b'-f(a)a'\]

Answer 25

Oh. Wow. Okay. I think I am starting to get it. Thank you so much amistre64!

Answer 26

youre welcome