Question

how can i evaluate... ill post it

Answer 1

\[y=\frac{d}{dx} \int\limits^{x^3}_{x} {\sin^2(t)}dt\]

using a double angle formula:

\[y=\frac{d}{dx} \int\limits^{x^3}_{x} {\frac{1}{2}(1 - \cos(2t))}dt\]

\[y=\frac{1}{2} \times \frac{d}{dx} ([x - \frac{1}{2} \sin2t]^{x^3}_x )\]

Answer 2

mistake, last line should be:

\[y=\frac{1}{2} \times \frac{d}{dx} ([t - \frac{1}{2} \sin2t]^{x^3}_x )\]

Answer 3

sub in the upper and lower limits, then differentiate the resulting mess, then divide by 2

Answer 4

first i resolve the parenthesis like an integral? then i look for its derivative and later i divide by 2?

Answer 5

yes

Answer 6

wait, the dividing by two is because we resolved it like an integral, are you ok with how i integrated?

Answer 7

i got lost in in the part that you integrated..

Answer 8

ok, ill go through it :)

Answer 9

lets forget about that d/dx for the moment, lets just consider this bit:

\[\int\limits^{x^3}_x{\sin^2t \text{ }dt}\]

Answer 10

the "t" here is what is known as a "dummy" variable (i think thats the word for it)

have you come across this identity before? :

\[\cos(2x) = \cos^2x - \sin^2x\]

Answer 11

yes

Answer 12

ok, so we want this, which we get from substituting (1-sin^2(x)) for cos^2(x) :

\[\cos(2x) = 1 - 2\sin^2x\]

\[\sin^2(x) = \frac{1}{2}(1-\cos(2x) )\]

Answer 13

now our integral becomes:

\[\int\limits^{x^3}_x{\frac{1}{2}(1-\cos(2x)) \text{ }dt}\]

Answer 14

ok, got it..

Answer 15

aaah sorry thats meant to be a t inside the cos
so we can take the 1/2 outside to worry about later

\[\frac{1}{2}\int\limits^{x^3}_x{1-\cos(2t) \text{ }dt}\]

Answer 16

do you follow?

Answer 17

yup

Answer 18

ok so can you integrate that?

Answer 19

no :s i get confused with those x's

Answer 20

ok, well forget about those for a second, can you see that
\[\int\limits1-\cos2t \text{ }dt = t -\frac{1}{2} \sin2t + C\]

Answer 21

yes

Answer 22

while that is the extremely long way to get to it; it will work only for integrations that can be integrated up to begin with; there is a simpler way ...

Answer 23

which one is the simplest way?

Answer 24

bah sorry..

Answer 25

ah the integral and differential cancel?

Answer 26

uisng the definitions for integration we get:

\[\int_{a}^{b}f(t)dt=F(b)-F(a)\]

correct?

Answer 27

and by the definition of derivative we get:

\[[F(b)-F(a)]'=f(b)b'-f(a)a'\]

Answer 28

soo\[\frac{d}{dx}\int_{a}^{b}f(t)dt=f(b)b'-f(a)a'\]

Answer 29

soooo
i use the equation that @eigenschmeigen gave me? (the one with the t-sin2t)

Answer 30

no .... not unless you want to finish up that route

but what im saying is that youve got all the informatiion you need al ready to determine the solution without doing any integration whatsoever

Answer 31

what is your f(t) ?

Answer 32

im not the one that needs to be able to recognize the f(t); i ask this for your benefit so that you can pick out the f(t) ...

Answer 33

\[\sin ^{2} (t)\]

Answer 34

good; what are our lower (a) and upper (b) limits of integration?

Answer 35

we are starting to fill in the answer with these things, so far we have

sin^2(b)b' - sin^2(a)a'

Answer 36

lower is x and upper is \[x^{3}\]

Answer 37

ok; lets place those in

\[sin^2(x^3)*[x^3]'-sin^2(x)*[x]'\]

what is the derivative of x^3; and the derivative of x ; with respect to x

Answer 38

derivative of x^3 is 3x^2

Answer 39

derivative of x is 1

Answer 40

then we have our final results

\[3x^2 sin^2(x^3)-sin^2(x)\]

Answer 41

thats the final answer?

Answer 42

theres nothing else to fill in ... my advice is, that when you get to the end , its over

\[\frac{d}{dx}\int_{a}^{b}f(t)dt=f(b)b'-f(a)a'\]

nothing left to work out now

Answer 43

thank you so much!

Answer 44

youre welcome