document.write( "Question 344815: Please help me solve this equation: \"+int%28+%281-x%29%5E2010%2C+dx%2C+0%2C+1+%29+\" \n" ); document.write( "
Algebra.Com's Answer #246707 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "I will put \"x=%22%22\" on the limits to show what variable the\r\n" );
document.write( "limits are on, namely x, since we will be changing the limits:\r\n" );
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document.write( "\"+int%28+%281-x%29%5E2010%2C+dx%2C+x=0%2C+x=1+%29+\"\r\n" );
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document.write( "Let \"u=1-x\"\r\n" );
document.write( "Then \"%28du%29%2F%28dx%29=-1\"\r\n" );
document.write( "or \"du=-dx\"\r\n" );
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document.write( "and \"dx=-du\"\r\n" );
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document.write( "So we substitute \"u\" for \"%281-x%29\" and \"-du\" for \"dx\"\r\n" );
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document.write( "\"+int%28+u%5E2010%2C+%28-du%29%2C+x=0%2C+x=1+%29+\"\r\n" );
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document.write( "And we can take out the negative sign in front of the integral:\r\n" );
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document.write( "\"+-int%28+u%5E2010%2C+du%2C+x=0%2C+x=1+%29+\"\r\n" );
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document.write( "But the limits are on x.  We want to change the limits\r\n" );
document.write( "so that they will be on u rather than x.\r\n" );
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document.write( "To do that we substitute the limits for x in the equation\r\n" );
document.write( "for u which is \"u=1-x\"\r\n" );
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document.write( "when \"x=0\", \"u=1-0=1\" and\r\n" );
document.write( "when \"x=1\", \"u=1-1=0\",\r\n" );
document.write( "so we change the limits to:\r\n" );
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document.write( "\"+-int%28+u%5E2010%2C+du%2C+u=1%2C+u=0+%29+\"\r\n" );
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document.write( "and we use the power formula \"int%28+u%5En%2C+du%29=u%5E%28n%2B1%29%2F%28n%2B1%29%2BC\"\r\n" );
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document.write( "Edwin

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