Question

intergals help.
i am confused

Answer 1

with FTOC problems

Answer 2

did you understand the last answer i wrote?
this one is similar, but you have to break up the integral in to two parts

Answer 3

could you show me the steps?

Answer 4

sure

Answer 5

if it was just
\[\int_0^x t^5dt\] then the derivative would be
\[x^5\] since the derivative of the integral is the "integrand"

Answer 6

if it was
\[\int_0^{x^4}t^5dx\] then you would have to use the chain rule to find the derivative because this is a composition of functions
the derivative would be
\[(x^4)^5\times 4x^3\]

Answer 7

in your case you have to break the integral in to two parts,
\[\int_x^{x^4}t^5dt\]
\[=\int _x^0t^5dt+\int _0^{x^4}t^5dt\]
\[=-\int _0^xt^5dt+\int_0^{x^4}t^5dt\]

Answer 8

then use the chain rule for each part

Answer 9

for both of the sides?

Answer 10

\[\large \int_{u(x)}^{v(x)}f(t)dt = F(v(x)) - F(u(x))\]


where F(t) is the antiderivative of f(t)

--------------------------------------


\[\large f(x) = \int_{u(x)}^{v(x)}f(t)dt\]


\[\large f^{\prime}(x) = \frac{d}{dx}\left(\int_{u(x)}^{v(x)}f(t)dt\right)\]


\[\large f^{\prime}(x) = \frac{d}{dx}\left(F(v(x)) - F(u(x))\right)\]


\[\large f^{\prime}(x) = \frac{d}{dx}\left(F(v(x))\right) - \frac{d}{dx}\left(F(u(x))\right)\]


\[\large f^{\prime}(x) = F^{\prime}(v(x))*v^{\prime}(x) - F^{\prime}(u(x))*u^{\prime}(x)\]

You use the chain rule on the last line above.

Answer 11

5t^4?

Answer 12

oh and forgot that F ' (t) = f(t)

so you can add in the line

\[\large f^{\prime}(x) = f(v(x))*v^{\prime}(x) - f(u(x))*u^{\prime}(x)\]

Answer 13

you have \[h(x)=\int_x^{x^2}t^5dt=\int_x^{x^2}f(t)dt\]
\[h'(x)=f(x^2)\times{d\over dx}(x^2)-f(x)\times{d\over dx}(x)\]
\[h'(x)=(x^2)^5\times(2x)-(x^5)\times(1)\]
simplify

Answer 14

h(x)'=x^10x(2x)-(x^5)*1?

Answer 15

good can simplify a little bit more

Answer 16

o, 2x^6?

Answer 17

@electrokid

Answer 18

no
\[x^{10}(2x)-x^5=\mathbf{2x^{11}-x^5}\]

Answer 19

ops, i didnt see - in front of x^5

Answer 20

this isnt final answer is it?

Answer 21

@electrokid