Question

Como eu resolveria essas duas questões?

Answer 1

A)

$\lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \\\\ \lim_{h \to 0} \frac{10(2+h)^2+(2+h)-42}{h}\\\\ \lim_{h \to 0} \frac{10(4+4h+h^2)+2+h-42}{h}\\\\ \lim_{h \to 0} \frac{40+40h+10h^2+2+h-42}{h}\\\\ \lim_{h \to 0} \frac{42+41h+10h^2-42}{h}\\\\\lim_{h \to 0} \frac{41h+10h^2}{h}\\\\\lim_{h \to 0} \frac{41h+10h^2}{h}\\\\\lim_{h \to 0} \frac{h(41+10h)}{h}\\\\\lim_{h \to 0}41+10h=41$

B)

$\int\limits^5_0(3t^2+30t+36)\ dt =\\\\\\3\int\limits^5_0 t^2 \ dt +30\int\limits^5_0 t \ dt +\int\limits^5_0 36 \ dt$

Agora vamos resolver separadamente:

$3\int\limits^5_0 t^2 \ dt =3(\frac{t^3}{3})]^5_0 = 3(\frac{5^3}{3}-\frac{0^3}{3} )=125\\\\ +30\int\limits^5_0 t \ dt =30(\frac{t^2}{2} )]^5_0=30(\frac{5^2}{2}-\frac{0^2}{2} )=375\\\\\int\limits^5_0 36 \ dt=36t]^5_0 =36\cdot5-36\cdot0=180$

Por fim temos:

$\int\limits^5_0 3t^2+30t +36 \ dt=\\3\int\limits^5_0 t^2 \ dt +30\int\limits^5_0 t \ dt +\int\limits^5_0 36 \ dt=\\\\125+375+180=680 ua$