SOLUTION: re. function f(x)=(x-2)(1-e^x) a) I have to find points where the function crosses the x axis ( I have found (2,0) & (0.0) which I hope are correct. b) I need to write an expres

Algebra ->  Trigonometry-basics -> SOLUTION: re. function f(x)=(x-2)(1-e^x) a) I have to find points where the function crosses the x axis ( I have found (2,0) & (0.0) which I hope are correct. b) I need to write an expres      Log On


   



Question 1039145: re. function f(x)=(x-2)(1-e^x)
a) I have to find points where the function crosses the x axis ( I have found (2,0) & (0.0) which I hope are correct.
b) I need to write an expression involving a definite integral that gives the area between the graph of f and the x axis from x=0 to x=2. Not sure how to write the symbol on here, but I think the answer is maybe the symbol f (without line across) with upper power of 2 and lower of 0 , dx
c) use integration by parts to find the area described in part (b) giving both the exact answer and an approximation to 3 decimal places. I'm really stuck with this so a step by step answer would be great.
Thank you for any help

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
The roots are correct.
int (x-2)(1-e^x)dx
let u=x-2
du=dx
dv=(1-e^x)dx
v=x-e^x
the integral is uv-int v*du
that is (x-2)(x-e^x)- int (x-e^x)dx; the integral is int x dx-int (e^x dx), so that is x^2/2-e^x
putting it together, it is x^2-xe^x-2x+2e^x-x^2/2+e^x, evaluated at 2 and 0.
at 2, it is 4-2e^2-4+2e^2-2+e^2. The fours cancel and the 2e^2 cancel to leave -2+e^2.
at 0, it is 0-0-0+2-0+1=3
subtracting -2+e^2-3=e^2-5 or 2.389
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