Question

How do you find the area of the region under the given curve y = (2x + 2)^1/2 on the interval [0,1]?

Answer 1

`8/3- 2sqrt2 /3`

Explanation:

The area of the region bounded by the curve and the x-axis would be `int_0^1 ydx` = `int_0^1 sqrt(2x+2) dx` = `2/3 sqrt2 (x+1)^(3/2)]_o^1`

=`2/3 sqrt2 (2^(3/2) -1)`= `8/3- 2sqrt2 /3`