What is #f(x) = int 3x-3secx dx# if #f((7pi)/4) = 0 #?

Question

What is #f(x) = int 3x-3secx   dx# if #f((7pi)/4) = 0  #?

Emma Aldridge · Accepted Answer

#f(x) = 3/2x^2 - 3ln|sec x + tan x| - 49.545#

I assume you mean #f(x) = int (3x - 3 secx)dx#.

So, We have,

#f(x) = int(3x - 3secx) dx#

#= 3int xdx - 3intsec xdx#

#= 3/2x^2 - 3ln|sec x + tan x| + C#

Now, According to the Question,

#color(white)(xxx)f((7pi)/4) = 0#

#rArr 3/2((7pi)/4)^2 - 3ln|sec ((7pi)/4) + tan ((7pi)/4)| + C = 0#

#rArr 3/2(22/4)^2 - 3ln|(-1/4) + 0| + C = 0#

#rArr 3/2 * 121/4 - 3(-1.39) + C = 0# [As #ln(1/4) = -1.39# (approx)]

#rArr 49.545 + C = 0# [Using Calculator]

#rArr C = -49.545#

So, #f(x) = 3/2x^2 - 3ln|sec x + tan x| - 49.545#

Hope this helps.

Samantha Adkison · Answer

undefined To find $ f(x) = \int (3x - 3\sec x) \, dx $ if $ f\left(\frac{7\pi}{4}\right) = 0 $, we first integrate $ 3x - 3\sec x $ with respect to $ x $:

$$ \int (3x - 3\sec x) \, dx = \frac{3}{2}x^2 - 3\ln|\sec x + \tan x| + C $$

Now, plug in $ x = \frac{7\pi}{4} $:

$$ f\left(\frac{7\pi}{4}\right) = \frac{3}{2}\left(\frac{7\pi}{4}\right)^2 - 3\ln|\sec\left(\frac{7\pi}{4}\right) + \tan\left(\frac{7\pi}{4}\right)| + C $$

Given that $ f\left(\frac{7\pi}{4}\right) = 0 $, we can solve for $ C $:

$$ 0 = \frac{3}{2}\left(\frac{49\pi^2}{16}\right) - 3\ln|-\sqrt{2} - 1| + C $$
$$ C = 3\ln|-\sqrt{2} - 1| - \frac{147\pi^2}{32} $$