Definite integral: Given 5x^3+40=int[a,x][f(t)]dt, find 'a'

dandaman

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Given: \(\displaystyle 5x^3+40=\int_{a}^{x}f(t)dt\) the value of a is?

my steps
-I am sorry but i don't even know how to start it. I was thinking about taking the derivative of both sides but I am not completely sure...
All help is appreciated.
 
Re: Definite integral question

start by taking the derivative of both sides of the equation w/r to x.
 
Re: Definite integral question

You are thinking along the right lines.

Here is a helpful rule from the second fundamental rule of calculus:

\(\displaystyle \frac{d}{dx}\int_{a}^{g(x)}f(t)dt=f(g(x))g'(x)\)
 
Re: Definite integral question

so i took the derivative of both sides and ended up with
\(\displaystyle 15x^2=f(t)\)
is this right?
if it is, would I then solve for x?
I am really not sure. Sorry about the clueless-ness but this problem has me stumped. :(
 
Re: Definite integral question

dandaman said:
so i took the derivative of both sides and ended up with

\(\displaystyle 15x^2=f(t)\) <<< How can that be?? 15x^2 is a function of 'x' - and f(t) should be function of 't'!!

is this right?
if it is, would I then solve for x?
I am really not sure. Sorry about the clueless-ness but this problem has me stumped. :(

Assuming you meant:

f(x) = 15 x^2

then

f(t) = 15 t^2

put this back into your original problem - (definite integral from 'a' to 'x' = 5x^3 + 40) - and solve for 'a'
 
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