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area of region bounded by graphs y=0, y=(1/9)xe^(-x/3),x=0,x=3?
I just can't get this one. Work would be appreciated
1 Answer
- Ossi GLv 71 decade agoFavorite Answer
Hello
Area = ∫(x/9*e^(-x/3))dx from 0 to 3
A = [-1/3*e^(-x/3)*(x+3)]
A = [- 1/3*e^-1(6)] - [- 1/3*e^0(3)]
A = - 0.73575888 + 1
A = 0.264241
Regards
Integration part by Wolfram alpha:
integral 1/9 e^(-x/3) x dx
Factor out constants:
= 1/9 integral e^(-x/3) x dx
For the integrand e^(-x/3) x, integrate by parts, integral f dg = f g- integral g df, where
f = x, dg = e^(-x/3) dx,
df = dx, g = -3 e^(-x/3):
= 1/3 integral e^(-x/3) dx-1/3 e^(-x/3) x
The integral of e^(-x/3) is -3 e^(-x/3):
= -1/3 e^(-x/3) x-e^(-x/3)+constant
Which is equal to:
= -1/3 e^(-x/3) (x+3)+constant
