Question

Calcule as integrais abaixo

Answer 1

a)

$I = \int\limits^{270\°}_{90\°} {sen(x)} \, dx$

$I=-cos(x)\limits|^{270\°}_{90\°}$

$I = -cos(270\°)-(-cos(90\°))$

$I = -cos(270\°)+cos(90\°)$

$I = 0+0$

$I = 0$

b)

$I = \int\limits^{4}_{0} {(e^x+x^3)} \, dx$

$I = \int\limits^{4}_{0} {e^x} \, dx + \int\limits^{4}_{0} {x^3} \, dx$

$I = e^{x}\limits|^{4}_{0}+\frac{x^4}{4}\limits|^{4}_{0}$

$I = (e^{4}-e^{0})+(\frac{4^4}{4}-\frac{0^4}{4})$

$I = (e^{4}-1)+(\frac{256}{4}-0)$

$I = e^4-1+64$

$I = e^4+63$

c)

$I = \int\limits^{4}_{1} {(100x^{-3}+\frac{1}{x})} \, dx$

$I = \int\limits^{4}_{1} {100x^{-3}} \, dx+ \int\limits^{4}_{1} {\frac{1}{x}} \, dx$

$I = (\frac{100}{-2}x^{-2})\limits|^{4}_{0}+(ln|x|)\limits|^{4}_{0}$

$I = (-50x^{-2})\limits|^{4}_{0}+(ln|x|)\limits|^{4}_{0}$

$I = -50(4^{-2}-1^{-2})+ln|4|-ln|1|$

$I = -50(\frac{1}{16}-1)+ln|4|-ln|1|$

$I = \frac{-50}{16}+50+ln|4|$

$I = -3,125+50+ln|4|$

$I = 46,875+ln|4|$

Portanto, letra D)