Find the area bounded by the following graphs

y = x^2 , y = 50 - x^2 for 0 (smaller than or equal) x (smaller than or equal) 9.
Thanks.

These graphs cross where
x² = 50 - x²
hence x = ±5

Presumably the vertical lines x = 0, x = 9 are also part of the boundary of the required area.
So we have to find the area between x = 0 and x = 5,
then between x = 5 and x = 9
and total them.

1 ∫ [(50-x²) - x²] dx from 0 to 5
= ∫ [50 - 2x²]dx
= [50x - 2x³/3] from 0 to 5
= (250 - 250/3) - 0
= 500/3

2. ∫ [x² - (50-x²)]dx from 5 to 9
= [2x³/3 - 50x] from 5 to 9
= (486 - 450) - (250/3 - 250)
= 608/3

Total area = 500/3 + 608/3
............ = 1108/3 = 369.3333333.... sq unit

First, Sketch a graph. 50-x^2 is above x^2 between the points of intersection. Then x^2 is above 50-x^2 for the rest of the graph. For this interval (0,9) we need the points of intersection.

50-x^2=x^2
50= 2x^2
25=x^2
X=+/-5

So integrate in two parts:

INT[50-x^2-x^2] dx on (0,5) + INT [ x^2-(50-x^2)] dx on (5,9)

= INT [ 50-2x^2] dx on (0,5) + INT [2x^2-50] dx on (5,9)

= [50x-2(x^3)/3] |(0,5) + [2(x^3)/3-50x] |(5,9)

= [(250-250/3) -( 0-0)] + [ (486-450) -( 250/3 -250)]

= 500/3 + (36+ 500/3)

= 369 1/3

Hoping this helps!