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calc definate integral from 0 to -5 of x times rt(4-x)
- Thread starter scotthen
- Start date
Re: calc definate integral
What is this supposed to be?.
Perhaps,
\(\displaystyle \L\\\int_{-5}^{0}x\sqrt{4-x}dx\)?.
scotthen said:how do I solve this equation...
integral from 0 to -5 of x times rt(4-x) ???[/tex][/url][/list][/list][/code]
What is this supposed to be?.
Perhaps,
\(\displaystyle \L\\\int_{-5}^{0}x\sqrt{4-x}dx\)?.
One way would be to use change of variables.
Let u = 4-x, then x = 4-u and dx=-du.
Your integral changes to:
\(\displaystyle \L \int_9^4 (4-u)u^{\frac{1}{2}}(-du)\,\, =\,\, - \int_9^4 (4u^{\frac{1}{2}}-u^{\frac{3}{2}})du = \int_4^9 (4u^{\frac{1}{2}}-u^{\frac{3}{2}})du\)
Note that the bounds change by the equation u=4-x, and I got rid of the negative by reversing the bounds.
Let u = 4-x, then x = 4-u and dx=-du.
Your integral changes to:
\(\displaystyle \L \int_9^4 (4-u)u^{\frac{1}{2}}(-du)\,\, =\,\, - \int_9^4 (4u^{\frac{1}{2}}-u^{\frac{3}{2}})du = \int_4^9 (4u^{\frac{1}{2}}-u^{\frac{3}{2}})du\)
Note that the bounds change by the equation u=4-x, and I got rid of the negative by reversing the bounds.
Another, slightly different substitution may be:
\(\displaystyle \L\\u^{2}=4-x \;\ and \;\ -2udu=dx \;\ and \;\ 4-u^{2}=x\)
Don't forget to change your limits of integration, accordingly.
\(\displaystyle \L\\2\int_{2}^{3}(4u^{2}-u^{4})du\)
\(\displaystyle \L\\u^{2}=4-x \;\ and \;\ -2udu=dx \;\ and \;\ 4-u^{2}=x\)
Don't forget to change your limits of integration, accordingly.
\(\displaystyle \L\\2\int_{2}^{3}(4u^{2}-u^{4})du\)