These two questions are ones that I am just stuck on...
#1. I found that
f(x) = -5x
but I need some help finding what
a = ?
Thanks for the help!
-qbkr21
This is the second question i've seen today that reminds me of the second fundamental theorem of calculus, but it's not in the exact form it needs to be. anyway, this is what i'd do:
since $\displaystyle f(x) = -5x$, $\displaystyle f(t) = -5t$
$\displaystyle \Rightarrow 1 + \int_{a}^{x} \frac {f(t)}{t^3}dt = 1 + \int_{a}^{x} \frac {-5t}{t^3}dt$
$\displaystyle \Rightarrow 1 - 5 \int_{a}^{x} \frac {1}{t^2}dt = -5x^{-1}$
$\displaystyle \Rightarrow \int_{a}^{x} \frac {1}{t^2}dt = \frac {1}{5} + x^{-1}$
$\displaystyle \Rightarrow -(x)^{-1} + (a)^{-1} = \frac {1}{5} + x^{-1}$
and solve for $\displaystyle a$
But this is speculation
Note that $\displaystyle f(x)$ is always a constant, it's value only depends on the value of $\displaystyle x$. $\displaystyle f(t)$ is no different.
$\displaystyle g(x) = \int_{-4}^{x} f(t)dt = [t]_{-4}^{x} = x - 4$
$\displaystyle \Rightarrow g(-6) = 0$, since $\displaystyle x<-4$
i think you have it from here
EDIT: o sorry, i think $\displaystyle g(-6) = -6 - 4 = -10$ i was treating $\displaystyle g(x)$ as $\displaystyle f(x)$