document.write( "Question 398002: The Question:
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Algebra.Com's Answer #282010 by richard1234(7193)\"\" \"About 
You can put this solution on YOUR website!
\"graph%28200%2C+200%2C+-5%2C+5%2C+-5%2C+5%2C+1+-+e%5E%28-2x%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "The area is normally given by \"int%28%281+-+e%5E%28-2x%29%29%2C+dx%2C+-1%2C+1%29\". However, part of the area is counted as negative, so we must put a negative wherever \"1+-+e%5E%28-2x%29+%3C+0\".\r
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\n" ); document.write( "\n" ); document.write( "The function goes below the x-axis when \"x+%3C+0\", so I'll split into two integrals:\r
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\n" ); document.write( "\n" ); document.write( " (There'll be lots of negative signs involved so I just rewrote the first integral)\r
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\n" ); document.write( "\n" ); document.write( "To evaluate the indefinite integral \"int%28%281-e%5E%28-2x%29%29%2C+dx%29\", begin with the substitution \"u+=+-2x\", \"du+=+-2dx\". Replacing dx, we obtain\r
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\n" ); document.write( "\n" ); document.write( "I'll let you finish off the problem by evaluating at the bounds (the rest is just arithmetic and applying the fundamental theorem of calculus).
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