Question

Calculus
For the following determine if the integral converges. If the integral converges, compute its value.
0∫inf (1+6x)^(-3/2)dx

Answer 1

\[\int\limits_{0}^{\infty} \left( \left( 1+6x \right)^{\frac{ -3 }{2 }} \right)dx=\frac{ 1 }{ 6 }\frac{ \left( 1+6x \right)^{\frac{ -1 }{ 2 }} }{ \frac{ -1 }{ 2 } } ~from~0~\to~\infty \]
solve it.

Answer 2

(-2/3)(1+6(0))^(-1/2)=0
(-2/3)(1+6(inf))^(-1/2)=>(-2/3)(sqrt(1))=>(-2/3)

0-(2/3)=(-2/3)?

Answer 3

\[=\frac{ -1 }{ 3 }\frac{ 1 }{ \sqrt{1+6x} }~from~0~\to~\infty \]
\[=\frac{ -1 }{ 3 }{\left[ \frac{ 1 }{ \sqrt{x}\sqrt{\frac{ 1 }{ x }+6} } \right]~x \rightarrow \infty}- \left( -\frac{ 1 }{ 3 } \right)\left( \frac{ 1 }{ \sqrt{1+6x} } \right)~x \rightarrow0\]
\[=-\frac{ 1 }{ 3 }\left[ 0-\frac{ 1 }{ \sqrt{1+0} } \right]=\frac{ 1 }{ 3 }\]

Answer 4

Thank you. But where did the 1/6 come from in the beginning?

Answer 5

\[\int\limits f^n(x)f'(x)dx=\frac{ f ^{n+1}(x) }{ n+1 }\]
here f(x)=1+6x,f'(x)=6,so i divide by 6 and multiply by 6