2006-10-11 05:25:43 · 4 answers · asked by richeboi 2 in Science & Mathematics ➔ Mathematics
That area is numerically equal to the integral of (1+ 2e^-x) over x ranging from 0 to 1,
area = Integral{ 1+ 2e^-x; 0
= 1- 2e^-1+ 2
= 3- 2e^-1.
2006-10-11 05:33:15 · answer #1 · answered by Illusional Self 6 · 0⤊ 0⤋
The answer is 3 - 2/e. This is equal to 2.26424112. You can find the answer by integrating 1+2e^-x from x = 0 to 1.
2006-10-11 05:32:57 · answer #2 · answered by Gabe 1 · 1⤊ 0⤋
You need to do the integral of 1+2e^-x from 0 to 1, and the answer is 3-2e^-1
2006-10-11 05:30:01 · answer #3 · answered by Kaylee 1 · 1⤊ 0⤋
§y(x)dx, from 0 to 1 = the enclosed area
§[1+2e^-x]dx, from 0 to 1
§1dx + 2§(e^-x)dx, from 0 to 1
[x + 2(-1)e^-x] from 0 to 1
[x - 2e^-x] from 0 to 1
[1-2e^-1] - [0 - 2e^0] = 3-2e^-1 = 2.26
2006-10-11 05:32:25 · answer #4 · answered by Mariko 4 · 1⤊ 0⤋