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Given f(x)=x^5+2x-1, what is `int_0^1` f(x)dx?
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You need to evaluate the definite integral, such that:
`int_0^1 f(x)dx = int_0^1 (x^5 + 2x - 1)dx`
Using the property of linearity of integral, you need to split the given integral in three simpler integrals, such that:
`int_0^1 f(x)dx = int_0^1 x^5 dx + int_0^1 2x dx - int_0^1 dx`
`int_0^1 f(x)dx = (x^6/6 + 2x^2/2 - x)|_0^1`
Reducing duplicate factors yields:
`int_0^1 f(x)dx = (x^6/6 + x^2 - x)|_0^1`
Using the fundamental theorem of calculus, yields:
`int_0^1 f(x)dx = (1^6/6 + 1^2 - 1 - 0^6/6 - 0^2 + 0)`
`int_0^1 f(x)dx = 1/6`
Hence, evaluating the given definite integral, yields `int_0^1 f(x)dx = 1/6.`
Posted by sciencesolve on September 30, 2013 at 5:14 PM (Answer #1)
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