Find the area of the region that is enclosed between y=x^2+1 and y=5 .

Question

I want solution

Find the area of the region that is enclosed between y=x²+1 and y=5

Answer 1

The region lies below the line y = 5 and above the graph of y = x² + 1.

 

The curves intersect when x²+1 = 5

 

                                            x² = 4    So, x = 2 or -2

 

Area of region = ∫_{(from -2 to 2)}(5 - (x²+1))dx

 

                     = 2∫_{(from 0 o 2)}(4 - x²)dx  (by symmetry)

 

                     = (8x - (2/3)x³)_{(from 0 to 2)} = 16 - 16/3 = 32/3