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First make a substitution and then use integration by parts to evaluate the integral.?

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First make a substitution and then use integration by parts to evaluate the integral.
∫ t^15 e^-t^8 dt

definite-integral

asked Apr 24, 2013 in CALCULUS by Jose Rodriguez Rookie

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2 Answers

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integral t to the power of 15 open parentheses e close parentheses to the power of minus t to the power of 8 end exponent d t equal integral t to the power of 8 cross times t to the power of 7 open parentheses e close parentheses to the power of minus t to the power of 8 end exponent d t

space space space space space space space space space space space space space space space space space space space space equal integral x cross times t to the power of 7 d t open parentheses e close parentheses to the power of minus x end exponent ( L e t space t to the power of 8 equal x space t h e n space 8 t to the power of 7 d t equal d x )

space space space space space space space space space space space space space space space space space space space space equal integral x cross times 1 over 8 d x open parentheses e close parentheses to the power of minus x end exponent (Substitute t to the power of 7 d t equal 1 over 8 d x )

space space space space space space space space space space space space space space space space space space space space equal 1 over 8 integral x cross times open parentheses e close parentheses to the power of minus x end exponent d x

( integral f open parentheses x close parentheses g open parentheses x close parentheses d x equal f open parentheses x close parentheses integral g open parentheses x close parentheses d x minus integral open parentheses d open parentheses f open parentheses x close parentheses close parentheses integral g open parentheses x close parentheses d x close parentheses d x right parenthesis

space space space space space space space space space space space space space space space space space space space space equal 1 over 8 open square brackets x open parentheses minus e to the power of minus x end exponent close parentheses minus integral minus e to the power of minus x end exponent d x close square brackets

space space space space space space space space space space space space space space space space space space space space equal 1 over 8 open square brackets x open parentheses minus e to the power of minus x end exponent close parentheses minus e to the power of minus x end exponent close square brackets

$space space space space space space space space space space space space space space space space space space space space equal fraction numerator minus e to the power of minus x end exponent over denominator 8 end fraction open square brackets x plus 1 close square brackets$

$space space space space space space space space space space space space space space space space space space space space equal fraction numerator minus e to the power of minus t to the power of 8 end exponent over denominator 8 end fraction open square brackets t to the power of 8 plus 1 close square brackets$ (Substitute x equal t to the power of 8 )

answered Apr 24, 2013 by diane Scholar

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Substitution method

Now the integral becomes

By parts of integration formula

Again substitute

answered Jul 9, 2014 by david Expert

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