Question

FTC part 2 ?

Answer 1

\[g(x)=\int\limits_{2}^{3x}(2t+3)dt\]

Answer 2

So do I just integrate and sub the limits as usual

Answer 3

everything turns into dx

Answer 4

my instructions are to find the derivative of each of the following functions defined by integrals

Answer 5

yeah and then
F(b) - F(a)

Answer 6

ok let me do it on my paper and can you double check my solution please

Answer 7

http://math.ucsd.edu/~wgarner/math20b/ftc.htm

Answer 8

yeah integration is not needed, just evaluate the bounds

Answer 9

so I just sub the limits and not integrate

Answer 10

\[\large g(x)=\int\limits_{2}^{3x}(2t+3)dt\]

\[\large g'(x)=\dfrac{d}{dx}\int\limits_{2}^{3x}(2t+3)dt\]

Answer 11

\[\large =(2(3x)+3) (3x)'\]

Answer 12

\[\large =(2(3x)+3) (3)\]

Answer 13

need an explanation ? :)

Answer 14

yes, please

Answer 15

lets work out a more general problem :

\[\large \dfrac{d}{dx} \int \limits_{a}^{g(x)} f(t) dt = ?\]

Answer 16

say, \(F(t) = \int f(t) dt\), then :

\[\int \limits_a^b f(t) dt = F(b) - F(a)\]

Answer 17

take the derivative with respect to x both sides :

\[\dfrac{d}{dx}\int \limits_a^b f(t) dt = \dfrac{d}{dx} F(b) - \dfrac{d}{dx}F(a)\]

Answer 18

In our present problem, b = g(x), a = some constant
since the derivative of constant is 0, we get :

\[\dfrac{d}{dx}\int \limits_a^b f(t) dt = \dfrac{d}{dx} F(b) - \dfrac{d}{dx}F(a)\]

\[ = \dfrac{d}{dx} F(b) - 0\]
\[ = \dfrac{d}{dx} F(b) \]

Answer 19

does that look okay so far

Answer 20

yes I am writing this all down on my class notes so I can study it. I have 6 of these types of problems to do.

Answer 21

next take the derivative using chain rule

Answer 22

keep in mind that b is not a constant number,
b = g(x) is a function of x

Answer 23

so my final solution is 18x+9

Answer 24

18x+9 is correct for the original problem

Answer 25

wait the second part for this one is d/dx F(2)

Answer 26

when I sub 2 I get 2(2)+3=7 and the derivative of a constant is zero

Answer 27

exactly ! so we don't need to bother about the lower bound

Answer 28

Formula :

\[\dfrac{d}{dx}\int \limits_a^{g(x)} f(t) dt = f(g(x)) \times g'(x)\]

Answer 29

if I have a number on either bound does that mean I disregard the upper or lower bound
In the next problem the upper bound is -1 so I will disregard the upper bound, correct

Answer 30

yep ! cuz when u differentiate it, it vanishes..

Answer 31

thanks, looks like now I know how to do the other 5 problems. Thank so much

Answer 32

careful about the signs...

\(\int_a^b \ne \int_b^a\)

Answer 33

\(\int_a^b = - \int_b^a\)

Answer 34

yes I will look out for that

Answer 35

just making sure :) when you have g(x) in the bottom, you need to change the formula accordingly - add a negative sign in front....

Answer 36

yes that is one of the properties of integrals or something like that (I don't recall it exact name)

Answer 37

Can you check the next one I did? the lower bond is -2

Answer 38

only if you have time, if not I am confident in your teaching and my learning.