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What is $f (x) = \int 3 x - 3 sec x d x$ $f(x) = int 3x-3secx dx$ if $f (\frac{7 π}{4}) = 0$ $f((7pi)/4) = 0$ ?

1 Answer

Soumalya Pramanik

May 20, 2018

$f (x) = \frac{3}{2} x^{2} - 3 ln | sec x + tan x | - 49.545$ $f(x) = 3/2x^2 - 3ln|sec x + tan x| - 49.545$

Explanation:

I assume you mean $f (x) = \int (3 x - 3 sec x) d x$ $f(x) = int (3x - 3 secx)dx$ .

So, We have,

$f (x) = \int (3 x - 3 sec x) d x$ $f(x) = int(3x - 3secx) dx$

$= 3 \int x d x - 3 \int sec x d x$ $= 3int xdx - 3intsec xdx$

$= \frac{3}{2} x^{2} - 3 ln | sec x + tan x | + C$ $= 3/2x^2 - 3ln|sec x + tan x| + C$

Now, According to the Question,

$\times x f (\frac{7 π}{4}) = 0$ $color(white)(xxx)f((7pi)/4) = 0$

$\Rightarrow \frac{3}{2} {(\frac{7 π}{4})}^{2} - 3 ln ∣ ∣ ∣ sec (\frac{7 π}{4}) + tan (\frac{7 π}{4}) ∣ ∣ ∣ + C = 0$ $rArr 3/2((7pi)/4)^2 - 3ln|sec ((7pi)/4) + tan ((7pi)/4)| + C = 0$

$\Rightarrow \frac{3}{2} {(\frac{22}{4})}^{2} - 3 ln ∣ ∣ ∣ (- \frac{1}{4}) + 0 ∣ ∣ ∣ + C = 0$ $rArr 3/2(22/4)^2 - 3ln|(-1/4) + 0| + C = 0$

$\Rightarrow \frac{3}{2} \cdot \frac{121}{4} - 3 (- 1.39) + C = 0$ $rArr 3/2 * 121/4 - 3(-1.39) + C = 0$ [As $ln (\frac{1}{4}) = - 1.39$ $ln(1/4) = -1.39$ (approx)]

$\Rightarrow 49.545 + C = 0$ $rArr 49.545 + C = 0$ [Using Calculator]

$\Rightarrow C = - 49.545$ $rArr C = -49.545$

So, $f (x) = \frac{3}{2} x^{2} - 3 ln | sec x + tan x | - 49.545$ $f(x) = 3/2x^2 - 3ln|sec x + tan x| - 49.545$

Hope this helps.

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