Asked by cathy
How do I solve for an area under the curve when y = x^2 and solve for above the interval (1,2) ? Any resources where I could see examples would be helpful. Thanks!
Answers
Answered by
Reiny
area = ∫ x^2 dx from x = 1 to 2
= [(1/3)x^3\ from 1 to 2
= (1/3)(8) - (1/3)(1)
= 7/3
google "area under curves" and pick a suitable page that you understand.
You will have to know basic Calculus, and basic integration.
= [(1/3)x^3\ from 1 to 2
= (1/3)(8) - (1/3)(1)
= 7/3
google "area under curves" and pick a suitable page that you understand.
You will have to know basic Calculus, and basic integration.