Question

Am nevoie doar de punctul c, mulțumesc ​

Answer 1

Răspuns:

Explicație pas cu pas:

c) f'  = 3x^3 -12x +12

3/(f' +12) = 1/(x^2 -4x+8) = 1/((x-2)^2 +2^2)

∫1/((x-2)^2 +2^2)dx = 1/2*arctg((x-2)/2)

Punand limitele:

1/2(arctg((4-2)/2) -arctg((2-2)/2)) =

1/2(pi/4 -0) = pi/8

Answer 2

$f:\mathbb{R}\rightarrow \mathbb{R},~~f(x)=x^3-6x^2+12x+5\\\\ \int\limits^2_4 \frac{3}{f'(x)+12}dx=\frac{\pi}{8} \\\\f'(x)=\Big(x^3-6x^2+12x+5\Big)'=3x^2-12x+12\\\\ f'(x)+12=3x^2-12x+12+12=3x^2-12x+24\\\\ \int\limits_2^4\frac{3}{3x^2-12x+24} dx=3\int\limits_2^4 \frac{1}{3(x^2-4x+8)} dx=\int\limits_2^4 \frac{1}{(x-2)^2+4}dx =\int\limits_2^4 \frac{1}{(x-2)^2+2^2} dx=$

$=\frac{1}{2} arctg\frac{x-2}{2} \Big|_2^4=\frac{1}{2} arctg\frac{4-2}{2} -\frac{1}{2} arctg\frac{2-2}{2} =\frac{1}{2} arctg\frac{2}{2} -\frac{1}{2} arctg\frac{0}{2} =\\\\=\frac{1}{2} arctg1-\frac{1}{2} arctg0=\frac{1}{2} \cdot \frac{\pi}{4} -\frac{1}{2} \cdot 0=\frac{\pi}{8} -0=\frac{\pi}{8}$