- #1
thomas49th
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Homework Statement
Find [tex]\int^{1}_{0} (x^{2} - 3x + 1)e^{x} dx[/tex]
Homework Equations
Let [tex]f =(x^{2} - 3x + 1)[/tex]
[tex g = e^{x}[/tex]
[tex] f' = 2x - 3[/tex]
[tex]\int (g) dx = e^{x}[/tex]
The Attempt at a Solution
We are going to have to use intergation by parts twice as the degree of the first function (f) is 2.
Important The question asks me to integrate between the limits 0 and 1. However is this asking me to find the area under the curve? I ask this because if I graph the functions' product, I am integrating above and below the x axis. Now if I was finding the area (which cannot be negative) it means I will have to integrate between the limits twice: 0 to 0.3458 and 0.3458 and 1, taking the modulus of the latter and adding them together. A glance at the question's answer shows the answer to be negative, telling me they weren't bothered about the finding the actual area, merely the summation of the area between the limits 0 and 1. Is there a SET RULE for knowing when to integrate in chunks depending on the x-axis to find the real world area and when to integrate without the worry of integrating between roots
Anyway back to my question. I am going to not bother to worry about the the graph crossing below the x axis, and integrate as a whole
First part of the integral:
[tex] (x^{2} - 3x + 1)e^{x} - \int^{1}_{0}(2x -3)e^{x}dx[/tex]
again,
[tex] (x^{2} - 3x + 1)e^{x} - (2x -3)e^{x} -2\int^{1}_{0}e^{x}dx[/tex]
gives
[tex] [(x^{2} - 3x + 1)e^{x} - (2x -3)e^{x} -2e^{x}]^{1}_{0}[/tex]
Putting the limits in yields
-2 -2e
when the answer is apparently -0.5634
Can anybody spot where I've gone wrong?
Thanks
Thomas