Question

Determine a área limitada pela curvas y = x^2 + x, y = x + 1 e pelas retas x = 2 e x = 4.

Answer 1

Considere a Integral definida em I = [2,4] de x² + x:

$\displaystyle \int^{4}_{2} x^{2}+x \, dx \\ \\ \\ \frac{1}{3}x^{3} + \frac{1}{2}x^{2} \left|\begin{array}{ccc}4\\\\\\2\end{array}\right \\ \\ \\ (\frac{1}{3}b^{3} + \frac{1}{2}b^{2})- (\frac{1}{3}a^{3} + \frac{1}{2}a^{2}) \\ \\ \\ (\frac{1}{3}4^{3} + \frac{1}{2}4^{2})- (\frac{1}{3}2^{3} + \frac{1}{2}2^{2}) \\ \\ \\ A= \frac{74}{3} \, u.a$

Agora considere a integral definida em I = [2,4] de x + 1:

$\displaystyle \int^{4}_{2} x + 1 \, dx \\ \\ \\ \frac{1}{2}x^{2} + x \left|\begin{array}{ccc}4\\\\\\2\end{array}\right \\ \\ \\ (\frac{1}{2}b^{2} + b)-(\frac{1}{2}a^{2} + a) \\ \\ \\ (\frac{1}{2}4^{2} + 4)-(\frac{1}{2}2^{2} + 2) \\ \\ \\ A = 8 \, u.a$

E a área entre as curvas é:

$\displaystyle A=\frac{74}{3} - 8 \\ \\ \\ \boxed{\boxed{A = \frac{50}{3} \, u.a}}$