Question

Find the area bounded by the curve y=x/1+x^2 and 4y=x.

Answer 1

curves intersect at points (0,0) and (-3/4, -3/16)

$\int^0_{-\frac 3 4}(-\frac {3x} 4 -x^2)dx=-\frac {x^3} 3 -\frac {3x^2} 8|^0_{-\frac 3 4}=-\frac{(-\frac 3 4)^3} 3 -\frac{3(-\frac 3 4)^2} 8=\frac 9 {128}=0.00703125$

Answer: 0.0703125