Question

help me to solve this question... Find the total area between the curve y=1-x^2, and the x-axis over the interval (0,2)...

Answer 1

you have to integrate the given curve's equation in the interval (0,2)
\[\int\limits_{0}^{2}1-x^2 = (2-0) + (-8/3 + 0) = 2 - 8/3 = -2/3\]
Since are cannot be negative, area = 2/3 sq.units

Answer 2

sorry I'm little confused while typing equations on OpenStudy..so I've been a little late in answering

Answer 3

oo its ok..the working just like that AbhimanyuPudi ?

Answer 4

may be u have to put another step in between..showing the integration..
\[\int\limits_{0}^{2}1-x^2 = x(from 0 \to 2) - x^3/3 (from 0 \to 2)\]

Answer 5

the negative sign in the answer indicates that the area is formed below the x-axis

Answer 6

for this case, first u have to figure out of the function y=1-x^2

so, the total area = A1 + A2