Question

The first part of the FTC states that if F(x) = integral of f(t) from a to x, then F'(x) = f(x). Using these facts and the properties of the definite integral find the derivative of ....ill write the integral using the equation tool

Answer 1

\[h(x) = \int\limits_{\sin^2x}^{x^2} tsin \sqrt{t} dt\]

Answer 2

Its hard to see but the bounds say: lower bound = sin^2 x and the upper bound is x^2

Answer 3

It is the derivative times the function according to FToC.

Answer 4

both the upper and lower boundry derivatives times f(g(x)) for both. Since you have two functions, you are basically doing it twice.

Answer 5

where g(x) is your boundries of integration functions.

Answer 6

So i separted the functions to \[\int\limits_{0}^{x^2}tsin \sqrt{t}dt - \int\limits_{0}^{si^2x}t \sin \sqrt{t}dt\]

Answer 7

So h(x) = f(g(x)) and \[f'(x) = xsin \sqrt{x}\]

Answer 8

If I remember correctly:
\[\int\limits_{0}^{v(x)}f(t)dt+\int\limits_{0}^{u(x)}f(t)dt\]

Answer 9

Ok so i did that part correctly. What should i do next

Answer 10

\[\frac{ d }{ dx }=-v'(x)f(v(x))+u'(x)f(u(x))\]

Answer 11

for the first integral, the v(x) should be below the zero.
\[\int\limits_{v(x)}^{0}\]

Answer 12

Oh so the lower bound should be x^2 and the upper bound is 0.

Answer 13

Yeah. That's why you get a -v'(x)

Answer 14

ok and now its \[-x^2(f(x^2)) + \sin^2x(f(\sin^2x))\]

Answer 15

Yeah.

Answer 16

well, that's not actually the derivative. so not exactly.

Answer 17

\[v'(x) \] and
\[u'(x)\]

Answer 18

ok so since v(x) = x^2 v'(x) = 2x.
u(x) = sin^2x so u'(x) = 2sinx(cosx)

Answer 19

yep.
but the original function is what gets plugged in as t.

Answer 20

you just multiply by the derivative.

Answer 21

so its -2x(f(x^2)) + 2sinxcosx ( f(sin^2 x))

Answer 22

yes. correct.

Answer 23

so we plug the x^2 and sin^2 x into the tsin*radical t equation??

Answer 24

yuup.

Answer 25

\[f(x) = x^2\sin \sqrt{x^2}\]

\[f(\sin^2x) = \sin^2x* \sin \sqrt{x^2} = sinx^3\sqrt{x^2}\]

Answer 26

for the 1st equation i mean f(x^2) not f(x)

Answer 27

You can't just combine both sin's because they are not the same. One is sin(x) and the other is sin(x^1/2)

Answer 28

and you also plug in sin^2(x) under the square root.

Answer 29

oh right its \[\sin^2x \sqrt{\sin^2x} = \sin^2x * sinx = \sin^3 x\]

Answer 30

so i cant combine the exponents for this one?

Answer 31

No. Both sin's are different.
However, if you had
\[\sin(2x)\sin(2x) = \sin^2(2x)\]

Answer 32

I hope this is correct now

Answer 33

\[-2x(x^2\sin \sqrt{x^2}) + 2sinxcosx( \sin^2x \sqrt{\sin^2x})\]

Answer 34

i should distribute this now right?

Answer 35

nothing cancels out so i guess i should simplify this as much as possible

Answer 36

yes. u should to make it look neater. but essentially, thats ur answer.

Answer 37

when u distribute the -2x to the first parenthesis do u get
\[-2x^3(-2xsin \sqrt{x^2})\]

Answer 38

that doesnt look right. What should it be

Answer 39

\[-2x^2(x^2\sin(\sqrt{x^2}+2\sin(x)\cos(x)\sin(\sqrt \sin^2(x))\]

Answer 40

the first term should be 2x. not 2x^2

Answer 41

why do u distribute the -2x to everything. I thought it was only for the 1st parenthesis

Answer 42

I didn't distribute the 2x to everything. I'm missing a parenthesis. here:
\[-2x(x^2\sin(\sqrt {x^2}) )+2\sin(x)\cos(x)(\sin^2(x)(\sqrt{\sin^2(x}))\]

Answer 43

ok yes. Thats also what i got. Its hard to simplify that :/

Answer 44

and you know by trig identities,
\[2\sin(x)\cos(x) = \sin(2x)\] if you wish to take it further. but I don't think you need to.

Answer 45

ok so u think thats enough simlification?

Answer 46

wait, nevermind. i think that trig identity is wrong. haha.

Answer 47

you might want to double check my trig identity.

Answer 48

i googled it now. U are correct

Answer 49

i am sure that for Calculus, that asnwer is good enough. simplifying further is algebra, and you are past that. it's assumed that you know your algebra tricks very well by now.

Answer 50

OK! THANK YOU 100 TIMES OVER! YOU are very patient to explain everything to me! Thanks bro. I really needed the help!

Answer 51

I understand teh calculus. Lord help me in algebra lol!

Answer 52

Calculus is just advance algebra. b/c as you may have noticed, it's the algebra that stoops most people. The calculus is relatively easy. You have formulas that u use to manipulate algebraically.

Answer 53

Yeah. The easy part is definitly the calculus part b/c its just understanding the concept. The algebra will just need some brushing up on

Answer 54

Thanks! This was more than enough help. Ur great man.

Answer 55

fa sho. best of luck, mate. Make sure you get integrals well cuz it gets way tougher in Calc-II.

Answer 56

Ok thanks for the tip! I'll make sure I get it down! Thanks. Peace.