y=x&2+9; [0,3]

Question

Find the area under the graph of the function over the interval given.

Answer 1

Function: y = x^2 + 9; [0, 3]

Find the area under the graph of the function over the interval given.

The area, A, is given by,

A = int[y] [0,3]

A = int[x^2 + 9] [0,3]

A = {x^3/3 + 9x} [0,3]

A = (27/3 + 27) - (0 + 0)

A = (9 + 27)

A = 36