Question

find the area bounded by the curve y=x^2 +2 and y= x + 2

Answer 1

Answer:

sorry don't know answer

Answer 2

Answer:

-1/6    

Step-by-step explanation:

y=x^2 + 2        y= x + 2

$x^{2}$ + 2 = x + 2

($x^{2}$ + 2 )- (x + 2)=0

$x^{2}$ - x = 0

x = 1 and x = 0

$\int\limits^0_1 {[(x + 2) - x^2 + 2]} \, dx = \int\limits^0_1 {(x - x^2)} \, dx$  

$=\frac{x^{2} }{2} + \frac{x^3}{3} \}_{1} ^{0}$

=$(\frac{1}{2} x^{2} - \frac{1}{3} x^{3} ) - (\frac{1}{2} x^{2} -\frac{1}{3} x^{3} )$

=$(\frac{1}{2} (0)^{2} - \frac{1}{3}(0)^{3} ) - (\frac{1}{2} (1)^{2} - \frac{1}{3} x^{3} )$

= 0 -$(\frac{1}{2} -\frac{1}{3} )$

=$-\frac{1}{2} + \frac{1}{3}$

= $-\frac{1}{6}$

not sure if that is correct . someone can check it back.