Question

find the area enclosed by the x-axis and the curve x=5+e^t y= t-t^2

Answer 1

range of t is?

Answer 2

it might be easier to define t in terms of x; then sub that into y

Answer 3

x=5+e^t

x-5 = e^t

ln|x-5| = t

Answer 4

Well, you could convert that to some awful function of x, which would then give you
\[\int_6^{5+e} f(x) dx\] but I'd much prefer to make the substitution into the variable t. Letting
\[x = 5 + e^t, dx = e^t dt\]
that gives
\[\int_0^1 (t-t^2)e^t dt = 3-e\]

Answer 5

y = t-t^2 roots on the xaxis at

0 = t(1-t); 0 and 1

ln|x-5| = 0, when x=5

ln|x-5| = 1, when x = e+5 ... such a nice number that is too

Jems looks easier by far, but I got no idea how to go about conceiving of it :)

Answer 6

6-5 = 1 ..... yeah

Answer 7

int: ln|x-5| = (x-5) (ln|x-5| - 1) if i recall the parttern correctly

Answer 8

In principle you can of course solve for t and plug it in to get a function of x. That would yield an integral
\[\int \ln(x-5)-\ln^2(x-5) dx\]
with bounds of six and 5+e corresponding to the points t=0 and t=1. All I'm saying is don't even bother making the substitution, because converting the above integrand back to a function of t just gives you
\[\int (t-t^2) e^t dt\]
with that extra exponential thrown in there because of the dx -> e^t dt substitution.

Answer 9

Plugging the first integral into wolfram alpha yields the same thing, but I surely wouldn't want to do it by hand :)

Answer 10

thank you everyone for your help :)