Answer to Question #156737 in Calculus for Sajal Abbasi

Question

• a) Sketch the region bounded by the graphs of the functions

                x²=x+1,   and    x= x²+4x+1.

• b) Determine the points of intersection of the curves given in part (a).
• c) Decide the limits of integration by using parts (a) and (b) and compute the area of the region that is bounded inside the graphs of both the curves. 

Answer 1

Here problem is wrongly typed

Correction made as

y²=x+1

y= x²+4x+1

a) Sketch is attached

b)

Solving y²=x+1 and

y= x²+4x+1 we get

x+1=(x²+4x+1)²

On simplification we get x=0 and x=-0.488

c)

Limits of integration are as follows

Lower limit is -0.488

Upper limit is 0

Area = $\int_{-0.488}^{0} [\sqrt{x+1}-(x²+4x+1)]dx$

= $[\frac{2}{3}(x+1)^{3/2}-\frac{x³}{3}-2x²-x]_{-0488}^{0}$

= 0.422 - 0.039 + 0.476 - 0.488

= 0.371 sq unit