Question

use the second fundamental theorem of calculus to find the derivative.

Answer 1

\[\int\limits_{4}^{2(x^2)+5?}t^3+1dt\]

Answer 2

\[\int\limits_{4}^{2(x^2) +5}t^3 +1 dt\] sorry this is what it should look like, no question mark

Answer 3

well i got taken to task for answering a question like this before, but no matter, i will answer again
first off the derivative of the integral is the integrand, so if you simply had
\[f(x)=\int_4^xt^3+1dt\] the derivative would be
\[f'(x)=x^3+1\]

Answer 4

but instead you have
\[f(2x^2+5)=\int_4^{2x^2+5}t^3+1dt\] so you have to find the derivative using the chain rule, as you would for any composite function
\[f(g(x))\]

Answer 5

replace \(t\) in the integrand by \(2x^2+5\) and then multiply by the derivative of \(2x^2+5\)

Answer 6

so the lower bound "4" has no bearing on finding the derivative?

Answer 7

no it is a constant.
if you had
\[\int_{2x}^{x^2}f(t)dt\] would be a different storyb

Answer 8

the lower limit of 4 determines the constant, and the derivative of a constant is zero

Answer 9

Ok that actually agrees with the other problems I have done. I guess I am just confused on the second fundemental theorem of calculus. The fundemental theorem is used to find the net change or posative area under a curve right? Then what is the purpose of the second fundemental theorem? My book doesn't explain it well

Answer 10

the "second fundamental theorem" says in english " the derivative of the integral is the integrand"

Answer 11

make sure to think of
\[\int_a^xf(t)dt\] as a function of \(x\) as in
\[F(x)=\int_a^x f(t)dt\] and so what the theorem tells you is that
\[F'(x)=f(x)\]

Answer 12

for example, if i had
\[\int_1^x2t^2+1dt\] then i know that if we put
\[F(x)=\int _1^x 2t^2+1dt\] then
\[F'(x)=2x^2+1\]

Answer 13

so it is really just a wy of taking an integral and finding its prime function?

Answer 14

I don't want to sound dumb but what does that apply to?

Answer 15

it also happens that we know another function with the same derivative, namely
\[G(x)=\frac{2}{3}x^3+x\] and therefore
\[F(x)=\int_1^x 2t^2+tdt=G(x)=\frac{2}{3}x^3+x+C\] this is what gives you the first fundamental theorem

Answer 16

what the theorem says is that if you have any reasonable ( integrable) function, that you can find its anti derivative, namely the integral

Answer 17

so you can use the second fundamental theorem to find the prime function, then integrate and arrive at the original function?

Answer 18

I didn't think of it like that. So its kind of like how you use logarithms to find exponetial functions and vise versa, this is a way to take an definite integal and revert to the original function?

Answer 19

you probably know that
\[\int_a^bf(t)dt=F(b)-F(a)\] where F is an anti derivative of \(f\) and the reason this is true is by the second fundamental theorem, so it should really be called the first one

Answer 20

so yes, since the derivative of the integral is the integrand it means if we can find ANOTHER such function (anti derivative) then it allows us an easy method to evaluate the integral, as in my example above

Answer 21

two functions have the same derivative so they differ by a constant.
i have to run but let me finish with the example above

Answer 22

ok

Answer 23

the two functions
\[\int_1^x2t^2+1dt\] and
\[\frac{2}{3}x^3+x\] both have the same derivative, namely
\[f(x)=2x^2+1\]

Answer 24

therefore we know that
\[G(x)=\frac{2}{3}x^2+x+C=\int_1^xt^2+1dt\]

Answer 25

the one on the right is the one we want to evaluate say we want
\[\int_1^2t^2+1dt\]
but it is hard to evaluate an integral, easy to evaluate a polynomial, so instead we evaluate
\[G(2)\]

Answer 26

we only have to solve for the constant, and since
\[\int _1^1t^2+1dt=0\] we know that \(G(1)=0\) and therefore
\(C=-(\frac{2}{3}+1)\)and hence the integral can be computed by
\[G(x)-G(1)\] as in first fundamental theorem

Answer 27

thanks for your help

Answer 28

yw