integrate xsqrt(4-x) dx by using integration by parts method.
\[\int x \times \sqrt{4-x} dx = x \int \sqrt{4-x} dx - \int [ \int \sqrt{4-x}dx \times \frac{dx}{dx}]\]
\[x \times \frac{2}{3} (4 -x)^{\frac{3}{2}} - \frac{2}{3} \times \frac{2}{5} \times (4-x)^{\frac{5}{2}}\]
the answers is \[2/3 (4-x)^{5/2} - 8/3 (4-x)^{3/2} +c\]
okay I am back let me do this again
ah you do remember the formula don't you
yeah i do i tried the question with v = (4-x)^1/2 and u = x
it gets me the wrong answer
i used u substitution by letting u= 4-x and i get the right answer but its not using the method that i want to lol
\[x \int \sqrt{4-x} dx - \int[ \int\sqrt{4-x}dx \times \frac{dx}{dx}]\]
ill brb :)
\[-x \times \frac{(4-x)^{1 + \frac{1}{2}}}{1 + \frac{1}{2}} - \int \frac{(4-x)^{1 + \frac{1}{2}}}{1 + \frac{1}{2}}dx\]
\[-x \times \frac{(4-x)^{\frac{3}{2}}}{\frac{3}{2}} - \frac{1}{\frac{3}{2}} \times \frac{(4-x)^{\frac{3}{2} + 1 }}{ 1 + \frac{3}{2}} \]
\[ -x \times \frac{2}{3} \times (4-x)^{\frac{3}{2}} - \frac{2}{3} \times \frac{2}{5} \times (4-x)^{\frac{5}{2}}\]
ah I did that again I am still getting the same with a sign reverse
ah Let's try this again with different approach
\[4- x = t \] \[-dx = dt\]
\[x = 4-t\]
\[\int ( 4- t )(-1)(\sqrt t) dt\]
\[\int (t - 4 ) \sqrt t dt \]
\[ \int t^{1 + \frac{1}{2}} - 4 \times t^{\frac{1}{2}}\]
\[\int t^{1 + \frac{1}{2}} - 4 \times t^{\frac{1}{2}}dt\]
\[\int t^{\frac{3}{2}}dt - 4 \int t^{\frac{1}{2}}dt\]
\[ \frac{t^{\frac{3}{2} +1 }}{\frac{3}{2} +1 } - 4 \frac{t^{\frac{1}{2} +1 }}{\frac{1}{2}+1}\]
\[\frac{2}{5} \times t^{\frac{5}{2}} - 4 \times \frac{2}{3} \times t^{\frac{3}{2}}\]
\[\frac{2}{5} \times (4-x)^{\frac{5}{2}} - \frac{8}{3} \times (4-x)^{\frac{3}{2}}\]
ah I am sure this is correct
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