Question

Find a function f and a number a such that
2+∫ a to x f(t)/t^4 dt = 6x^−1

f(x) =
a =

Answer 1

\[2+\int\limits_a^x \frac{f(t)}{t^4} dt=6x^{-1}\]
i think I would try differentiating both sides first

Answer 2

you will first be able to find f(x) this way
or f(t) if you replace the x's with t's later

then you will plug back and and solve the equation for a

Answer 3

if you want you can let me know what you have after differentiating both sides

Answer 4

well the derivative for the right side is -6x^-2. i'm not sure how to differentiate the left side :(

Answer 5

fundamental theorem of calculus
you did it last problem

Answer 6

derivative of 2 is 0 of course
now look at the integral and differentiate that part

Answer 7

f(x)/ x^4 or is there something more?

Answer 8

\[\frac{d}{dx} \int\limits _a^xg(t) dt=g(x)\] this one
yes! !!! that is it

Answer 9

solve for f(x) now

Answer 10

you have
\[\frac{f(x)}{x^4}=-6x^{-2}\]

Answer 11

-6x^2

Answer 12

beautiful

Answer 13

:)

Answer 14

now plug f(t)=-6t^2 in and solve for a
\[2+\int\limits\limits_a^x \frac{-6t^2}{t^4} dt=6x^{-1}\]

Answer 15

let me know if you need more help or if you are confused

Answer 16

yeah i need help. can you show me the process?

Answer 17

you can simplify the integrand then use power rule to integrate
then plug in limits

Answer 18

\[-6 \frac{t^2}{t^4}=-6t^{-2}\]
is integrand after simplification

Answer 19

integrate this integrand by use of power rule (power rule for integrals that is since this is an integral :p)

Answer 20

i'm kind of new to integrals, can you show mw what to do?

Answer 21

\[2+\int\limits\limits\limits_a^x \frac{-6t^2}{t^4} dt=6x^{-1} \\ 2+\int\limits_a^x -6t^{-2} dt=6x^{-1} \\ \text{ you are evaluating the integral right now }\]

recall
\[n \neq -1 \\ \int\limits_{a}^{b}x^{n} dx=\frac{x^{n+1}}{n+1}|_a^b=\frac{a^{n+1}}{n+1}-\frac{b^{n+1}}{n+1}\]

Answer 22

oops I pluggin in a first didn't mean to do that

Answer 23

\[n \neq -1 \\ \int\limits\limits_{a}^{b}x^{n} dx=\frac{x^{n+1}}{n+1}|_a^b=\frac{b^{n+1}}{n+1}-\frac{a^{n+1}}{n+1}\]

Answer 24

@lawls you okay?

Answer 25

yeah i just went somewhere sorry to keep you waiting

Answer 26

here is a big hint if you return:
you might end up with something like this after doing the evaluation of the definite integral:
\[2+g(x)-g(a)=g(x) \]

Answer 27

oops I was typing that before I noticed you came back lol

Answer 28

lol ok let me try to read what you wrote

Answer 29

i got [a^-1 / -1] - [x^-1/-1]

Answer 30

hmm do you mean this:
\[2+-6 \int\limits_a^x t^2 =6x^{-1} \\ 2+6 \int\limits_x^a t^2 =6x^{-1} \\ 2+6[-t^{-1}]_x^a=6x^{-1} \\ 2+6(-a^{-1}+x^{-1})=6x^{-1}\]
if so yes that is the right track

Answer 31

you can distribute the 6 there

Answer 32

and cancel out the functions of x's
just like we have 2+g(x)-g(a)=g(x)
we can cancel the g(x) by subtracting g(x) on both sides
2-g(a)=0

Answer 33

so you will have an equation just in terms of a

Answer 34

ok a =3

Answer 35

yeah that is what I have too :)
gj

Answer 36

thank you so much!!

Answer 37

np